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This answer is essentially user image's answer using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on proper time, $$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \cr g_{\mu\nu}~=~&{\rm diag}(-1,1,1,1) \end{align}\tag{1}$$ up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$

3. Then the Lagrangian reads $$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, \cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless.

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is $$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is $$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr ~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$ and the energy is $$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}, \end{align}\tag{7}$$ which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to Noether's theorem, the total momentum (energy) is conserved, respectively.

This answer is essentially user image's answer using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on proper time, $$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \cr g_{\mu\nu}~=~&{\rm diag}(-1,1,1,1) \end{align}\tag{1}$$ up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$

3. Then the Lagrangian reads $$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, \cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless.

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is $$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is $$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr ~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$ and the energy is $$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~\frac{m_0c^2}{\sqrt{1-\frac{{\bf v}^2}{c^2}}}, \end{align}\tag{7}$$ which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to Noether's theorem, the total momentum (energy) is conserved, respectively.

This answer is essentially user image's answer using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on proper time, $$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\mu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \cr g_{\mu\nu}~=~&{\rm diag}(-1,1,1,1) \end{align}\tag{1}$$ up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$

3. Then the Lagrangian reads $$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, \cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless.

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is $$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is $$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr ~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$ and the energy is $$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}, \end{align}\tag{7}$$ which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to Noether's theorem, the total momentum (energy) is conserved, respectively.

This answer is essentially user image's answer using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on proper time, $$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \cr g_{\mu\nu}~=~&{\rm diag}(-1,1,1,1) \end{align}\tag{1}$$ up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$

3. Then the Lagrangian reads $$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, \cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless.

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is $$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is $$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr ~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$ and the energy is $$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}, \end{align}\tag{7}$$ which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to Noether's theorem, the total momentum (energy) is conserved, respectively.

This answer is essentially user image's answer using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on proper time, $$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\mu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \end{align}\tag{1}$$ up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$

3. Then the Lagrangian reads $$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, \cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless.

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is $$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is $$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr ~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$ and the energy is $$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}, \end{align}\tag{7}$$ which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to Noether's theorem, the total momentum (energy) is conserved, respectively.

This answer is essentially user image's answer using slightly different words:

1. Argue that a Lorentz-invariant action principle for a massive point particle should be based on proper time, $$\begin{align} S[x] ~=~&\int\! d\lambda~L, \cr L~=~&\alpha \sqrt{-g_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\mu}}, \cr \dot{x}^{\mu}~:=~&\frac{dx^{\mu}}{d\lambda}, \cr g_{\mu\nu}~=~&{\rm diag}(-1,1,1,1) \end{align}\tag{1}$$ up to a proportionality factor $\alpha$.

2. Go to static gauge $$\lambda~=~t~=~\frac{x^0}{c}.\tag{2}$$

3. Then the Lagrangian reads $$\begin{align} L~\stackrel{(1)+(2)}{=}&~\alpha\sqrt{c^2-{\bf v}^2}, \cr {\bf v}~:=~&\dot{\bf x}.\end{align}\tag{3}$$

4. Argue that because the Lagrangian (3) should have dimension of energy, the constant $\frac{\alpha}{m_0c}$ should be dimensionless.

5. Argue that to recover the usual non-relativistic limit $|{\bf v}| \ll c$ for the Lagrangian $L$ [up to an irrelevant constant term that cannot be seen in the Lagrange equation, and which turns out to be the rest energy $E_0=m_0c^2$], the constant $$\frac{\alpha}{m_0c}~=~-1\tag{4}$$ must be minus one.

6. So we have derived that the Lagrangian is $$L~\stackrel{(3)+(4)}{=}~-m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}.\tag{5}$$

7. Therefore the canonical/conjugate momentum is $$\begin{align}{\bf p}~:=~&\frac{\partial L}{\partial {\bf v}}\cr ~\stackrel{(5)}{=}~&\frac{m_0{\bf v}}{\sqrt{1-\frac{{\bf v}^2}{c^2}}},\end{align}\tag{6}$$ and the energy is $$\begin{align} E~:=~&{\bf p}\cdot {\bf v}-L\cr ~\stackrel{(5)+(6)}{=}&~m_0c^2 \sqrt{1-\frac{{\bf v}^2}{c^2}}, \end{align}\tag{7}$$ which answer OP's question.

8. Furthermore, instead of just 1 point particle, we can generalize to a Lagrangian $L=\sum_{i=1}^N L_i$ for $N$ point particles, and we can even introduce interaction and potential terms. If the Lagrangian is invariant under space (explicit time) translations, then according to Noether's theorem, the total momentum (energy) is conserved, respectively.