When quantizing the electromagnetic quadri-potential, one makes the following ansatz :
\begin{equation*}
A_\mu (x^\sigma)= \int \frac{d^3 p}{(2\pi)^3 2E_p} \sum_{n=1}^2 e^{(n)}_\mu(p)\left(a_{(n)} (k) e^{-i k_\rho x^\rho}+a^\dagger_{(n)}(k)e^{ik_\rho x^\rho} \right)
\end{equation*}
Where $a^\dagger_{(n)}(k)$ is the creation operator of excitation of the quadri-potential field. By analyzing the symmetries of the field with the Noether theorem one comes to the conclusion that one has a particle with spin 1, no charge of any kind (so not a gluon) and $E_p=|\vec{p}|$. In short, one has identified a photon. One other way to answer is to be blind about all things discovered in physics until quantum field theory and say that the quadri-potential has excitations and that we call these excitations "photons".
Edit for the Noether currents:
First things first let us recall the Noether currents associated with the following transformation for $n$ scalar fields. Then I will give the Noether currents for the Proca lagrangian and then take the limit $m \rightarrow 0$. Let be the following transformations with $\alpha^j \ll 1$:
\begin{align*}
x^\mu & \longrightarrow x^\mu-f^\mu_j \alpha^j \\
\phi_i & \longrightarrow \phi_i+C_{ij} \alpha^j.
\end{align*}
Then the Noether currents of $\mathcal{L}[\phi_i]$ are:
\begin{equation*}
\Theta^\mu_j = \frac{\partial \mathcal{L}}{\partial \partial_\mu \phi_i}\left( C_{ij}+\partial_\nu \phi_i f^\nu_j\right)-f^\mu_j \mathcal{L}
\end{equation*}
The generalization to vectorial fields is straightforward since they are just four fields. For translation in space-time, we know that the Noether currents are respectively momentum and energy. So the transformations are $C_{ij}=0$ and $f^\mu_j \equiv f^\mu_{\nu}=\delta^\mu_\nu$. Then we have the following Noether currents :
\begin{equation*}
\Theta^\mu_0=\frac{\partial \mathcal{L}}{\partial \partial_\mu A_\nu} \partial_\rho A_\nu \delta^\rho_0-\delta^\mu_0 \mathcal{L} \,\,\,\,\,,\,\,\,\, \mathcal{L}=-\frac{1}{4}F^{\mu \nu}F_{\mu \nu}+\frac{1}{2}m^2 A_\mu A^\mu.
\end{equation*}
Slitting the time and spatial components and using the Proca equation one finds:
\begin{equation*}
\Theta_{00}=\frac{1}{2m^2}(\partial_i F_{0i})^2+\frac{1}{2}(F_{0i})^2+\frac{1}{4}(F_{ij})^2+\frac{1}{2}m^2 A_i^2.
\end{equation*}
Then using the ansatz I gave first in this answer and introducing the conjugate field $B^\mu=-F^{0\mu}$ for the quantization, and a lot of calculations (including a Bogolyubov transformation for the third creation operator) one has, after a shift in the vacuum energy:
\begin{equation*}
E=\int d^3x \Theta_{00}=\int d^3k\,E_\vec{k} \sum_n a^\dagger_{(n)}(\vec{k})a_{(n)}(\vec{k})\,\,\,\,\,,\,\,\,\,\,E_\vec{k}=\sqrt{|\vec{k}|^2+m^2}.
\end{equation*}
Then, acting on a ket $| 1_{\vec{p},n} \rangle$ of the state of one particle with momentum $\vec{p}$ and polarization $n$, the energy operator gives $E| 1_{\vec{p},n} \rangle=E_\vec{p}| 1_{\vec{p},n} \rangle$. Taking the limit $m\rightarrow 0$ one recovers the expression $E=|\vec{p}|$.
Now, for the spin we have the following transformation of the field, an infinitesimal rotation on the $(xy)$ plane:
\begin{align*}
x^1 & \longrightarrow x^1+\theta x^2\\
x^2 & \longrightarrow x^2-\theta x^1\\
A^1 & \longrightarrow A^1+\theta A^2\\
A^2 & \longrightarrow A^2-\theta A^1
\end{align*}
Then, $f^1_3=-x^2$, $f^2_3=x^1$, $C^1_3=A^2$ and $C^2_3=-A^1$. Using the formula I gave above for the Noether currents and putting $\mu=0$ in $\Theta^\mu_3$ to have the angular momentum, and recalling that $p^i=B_j \partial^i A^j$ we find:
\begin{equation*}
\Theta^0_3 = x^1 p^2-x^2 p^1+A^1 B^2-A^2 B^1.
\end{equation*}
Then one can do so for the other possible rotations and find that the total angular momentum is $\vec{M}=\vec{L}+\vec{S}$, where $\vec{S}=\int d^3x \vec{A} \wedge \vec{B}$. We have the property $[S_i,A_j]=i\epsilon_{ijk}A_k$, that give rise to the following one:
\begin{equation*}
[S_i,a^\dagger_{(n)}]=i\epsilon_{inm}a^\dagger_{(m)}
\end{equation*}
Then, one can easily calculate $[\vec{S}^2,a^\dagger_{(n)}]=i\epsilon_{inm}S_i A^\dagger_{(m)}+[S_i,a^\dagger_{(n)}]S_i$. By acting this quantity on the vacuum state we find:
\begin{align*}
\vec{S}^2 a^\dagger_{(n)} |0 \rangle &= i\epsilon_{inm}S_i a^\dagger_{(m)} |0\rangle\\
&=i\epsilon_{inm}S_i a^\dagger_{(m)} |0\rangle-\underbrace{i\epsilon_{inm}a^\dagger_{(m)} S_i |0\rangle}_{=0}\\
&= 2\delta_{nl} a^\dagger_{(l)} |0 \rangle
\end{align*}
But we know by the representation theory of the $\text{Spin}(3)$ group that $\vec{S}^2|\psi\rangle=j(j+1)|\psi\rangle$. Then one has $j=1$ for our field. In fact, one can actually find directly the spin from the representation theory of the $\text{Spin}(3)$ group.
Finally and this will be easy, the charge of the field. Since our field does not transform under the transformation $e^{-i\theta}A$, our field has no electric charge, and since it does not belong to any representation of group, it doesn't have any charge of any kind.
Hope this helped a bit, the calculations are pretty straightforward.