A general proof applying to reparametrization-invariant actions $$S = \int d^2 u \mathcal{L}(x^{\mu},x_{,i}^{\mu}), \ \ \ \ i = 0,1, \ \ \ \mu = 0,1,\dots, D- 1,$$ is to note that part of the derivation of Noether's theorem (the derivation where you assume the volume element varies also), schematically derived as follows (see the reference below for details) for an infinitesimal variation of the parameters $\tilde{u}^i(u) = u^i + \varepsilon^i(u)$ by noting \begin{align} \delta S &= \delta \int d^2 u \mathcal{L}(x^{\mu},x_{,i}^{\mu}) \\ &= \int d^2 u [\dfrac{ \partial \mathcal{L}}{\partial x^{\mu}} \delta x^{\mu} + \dfrac{ \partial \mathcal{L}}{\partial x^{\mu}_{,i}} \delta x^{\mu}_{,i} + \mathcal{L} \partial_i \varepsilon^i ] \\ &= \int d^2 u \{ \partial_i [ (\mathcal{L} \delta^i_j - \dfrac{ \partial \mathcal{L}}{\partial x^{\mu}_{,i}} x^{\mu}_{j})\varepsilon^j ] - ( \dfrac{\partial \mathcal{L}}{\partial x^{\mu}} - \dfrac{\partial }{\partial u^i} \dfrac{\partial \mathcal{L}}{\partial x_{,i}^{\mu}}) x^{\mu}_{,i} \} \\ &= 0, \end{align} implies, for variations of the $u^i$ which vanish at the boundaries, that the operators \begin{align} L_{\mu}(\partial x, \partial^2 x) = \dfrac{\partial \mathcal{L}}{\partial x^{\mu}} - \dfrac{\partial }{\partial u^i} \dfrac{\partial \mathcal{L}}{\partial x_{,i}^{\mu}} \end{align} satisfy the identities \begin{align} L_{\mu}(\partial x, \partial^2 x) x^{\mu}_{,i} = 0 , \ \ \ i = 0,1 \end{align} automatically, which means we only have $D - 2$ equations of motion for $D$ coordinates, so that the solution depends on two arbitrary functions of the parameters $u^0,u^1$.
For actions like the Nambu-Goto action not depending on the $x^{\mu}$ the operators $L_{\mu}$ take the form $$L_{\mu}(\partial x, \partial^2 x) = \dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}}{\partial x'^{\mu}}) $$ and satisfy the identities \begin{align} L_{\mu}(\partial x, \partial^2 x) \dot{x}^{\mu} &= 0 \\ L_{\mu}(\partial x, \partial^2 x)x'^{\mu} &= 0 \end{align} which when expanded become, say, for $\dot{x}^{\mu}$: \begin{align} 0 &= L_{\mu}(\partial x, \partial^2 x) \dot{x}^{\mu} \\ &= [\dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}}{\partial x'^{\mu}})]\dot{x}^{\mu} \\ &= [\dfrac{\partial}{\partial \tau} (\dfrac{\partial \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\mu}}) + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}(\dot{x},x')}{\partial x'^{\mu}})]\dot{x}^{\mu} \\ &= [ \dfrac{\partial^2 \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\nu} \partial \dot{x}^{\mu}} \ddot{x}^{\nu} + \dots + \dfrac{\partial}{\partial \sigma} (\dfrac{\partial \mathcal{L}(\dot{x},x')}{\partial x'^{\mu}})]\dot{x}^{\mu} \\ &= [\dfrac{\partial^2 \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\nu} \partial \dot{x}^{\mu}} \dot{x}^{\mu} ] \ddot{x}^{\nu} + \dots \end{align} and since this must hold regardless of the choice of $x^{\mu}$, the coefficients of each derivative of $x^{\nu}$ must be zero automatically, implying the Hessian satisfies $$[\dfrac{\partial^2 \mathcal{L}(\dot{x},x')}{\partial \dot{x}^{\nu} \partial \dot{x}^{\mu}} \dot{x}^{\mu} ] = 0$$ and similarly starting from $x'^{\mu}$, giving the rank of the Hessian to be $D- 2$. Furthermore analyzing the components of $$t_{ij} = \mathcal{L} \delta^i_j - \dfrac{ \partial \mathcal{L}}{\partial x^{\mu}_{,i}} x^{\mu}_{j} = 0$$ you get both primary constraints for the Nambu-Goto action. This is all done in:
Reference:
- Barbashov, B. M., & Nesterenko, V. V. (1990). Introduction to the relativistic string theory; Sec. 3, 7, Appendix B.