Why are photons excitations of Electromagnetic Fields? How do you know that the excitations of the Electromagnetic Field are photons? In the canonical quantization of the Electromagnetic Field, it could be derived that:
$$E_{p}=\left|p⃗\right|$$
However, this, at least by my intuition, doesn't make the excitations photons. You would still need to consider about the spin of the virtual particles (the charge, to my knowledge is determined to be zero, since it is a real vector field theory), and even then how would it be a photon and not something like a gluon? (is it because gluons don't interact with EM fields?)
I am not learning QFT professionally, so this may seem like a pretty naïve question.
 A: 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.
A: 
How do you know that the excitations of the Electromagnetic Field are photons

The word "field" in quantum field theory does not have  the same definition as in classical electromagnetic theory:

Moreover, within each category (scalar, vector, tensor), a field can be either a classical field or a quantum field, depending on whether it is characterized by numbers or quantum operators respectively.

As shown in innumerable experiments (example) it is the electromagnetic waves, light, that are  composed of photons.
One  starts from  the data to be modeled   when making a quantum field theory to fit the data:
One has the elementary particles in the table , electrons, neutrinos, photons, quarks etc, and makes the QFT by postulating a field represented by the plane wave solution of the corresponding   wave equation for each  particle  on which creation and annihilation operators work ( as shown in the other answer) to create and annihilate  the particles, in this case the photons.
Since the photon field is designed in order to model the created photons, the spin etc are mathematically imposed by the choice of the QFT.
