Here's a stab at a short and intuitive answer (i.e. math lite). Warning, I ignore basically all factors that aren't equal to 1 so sorry if I get something wrong. $\hbar = \pi = 1/2 = 2 = 4 = 1...$ etc.
Solutions to Maxwell's equations can be decomposed into monochromatic electromagnetic waves*. An electromagnetic wave has an (complex) amplitude $a\propto E$ which oscillates in time.
It can be shown that the energy (or energy flux depending on the exact physical situation) in the electromagnetic wave is proportional to the squared amplitude of the wave:
$$
H = \omega a^*a
$$
Instead of expressing the amplitude as a single complex number we can express it in terms of the real an imaginary parts of that number:
$$
a = \text{Re}(a) + i \text{Im}(a) = x + i p
$$
$$
H/\omega = x^2 + p^2
$$
WERE this a mass on a spring harmonic oscillator we would identify $x$ and $p$ as the canonically conjugate position and momentum variables for the classical harmonic oscillator. These are known to have a Poission bracket like
$$
\{x, p\} = 1
$$
Canonical quantization works by elevating the canonically conjugate variables to quantum operators (can be thought of as operators on a Hilbert space) with commutation relations:
$$
[\hat{x}, \hat{p}] = i
$$
It then follows that
$$
[\hat{a}, \hat{a}^{\dagger}] = 1
$$
from which one can derive that
$$
H = \omega\hat{a}^{\dagger} \hat{a} = \omega\hat{n}
$$
Where the eigenvalues of $\hat{n}$ are non-negative integers. This is what is meant by the energy of the harmonic oscillator being quantized. Note that this implies that the complex amplitude of the harmonic oscillator is quantized as well.
Back to the electromagnetic field. It turns out** that 1) electromagnetism can be cast in a Lagrangian form, thus giving us a context in which we can discuss canonically conjugate variables and 2) The real and imaginary parts of the electromagnetic field amplitude, $a$, given by $x$ and $p$, can in fact be related to a pair of conjugate variables of electromagnetism. These conjugate variables are related to the electric field and the time derivative of the electric field. There are also relationships, through Maxwell's equations, to the magnetic field.
This means that we are justified in making the theory of electromagnetism quantum by letting $a\rightarrow \hat{a}$ with
$$
[\hat{a},\hat{a}^{\dagger}]=1
$$
This leads us to the conclusion that the energy in a single mode of the electromagnetic is quantized in units of $\omega$, and likewise the amplitude of the electromagnetic wave is also quantized.
This is what a photon is. A photon is a quantized excitation in the electomagnetic wave, in the same way we can have a quantized excitation in the quantum harmonic oscillator. Also, at this point I'll point out that it is possible to have excitations in the classical electromagnetic field as well. If you ever find yourself feeling funny about photons try to spend more time thinking about the similarities, rather than the differences, between classical excitations of the electromagnetic field (or any oscillator) and quantum excitations of the same.
As described in the comments, the photon arises from electromagnetism through an application of canonical quantization to the canonically conjugate variables which make up a single-mode monochromatic solution to Maxwell's equations.
*These are often taken to be plane waves, but, depending on the boundary conditions for the EM problem, we can decompose the general solution to Maxwell's equations into many different families of spatial modes. Any such decomposition suffices for the description of a photon. The shape of the photon is given by the spatial pattern of the spatial mode that we are quantizing. That is photons can have different shapes, and a photon in one mode can be decomposed as a superposition of photons in other modes. There, I answered about 100 physics stack exchange questions about the shape of a photon in this footnote!
**My favorite references for this are Quantum and Atom Optics by Daniel Steck and UC Berkeley Physics 221A/B Lecture Notes by Robert Littlejohn.
Best Answer
The following is not well-known, but (modified) Maxwell equations can indeed describe both electromagnetic field and electrons.
@Quantumwhisp commented: "Maxwell's equations don't describe charged particles at all", and then asked: "Can you derive the Lorentz-Force from maxwell's equations?"
I am not saying these comments are unreasonable, but, surprisingly, Dirac did derive the Lorentz force from Maxwell equations (Proc. Roy. Soc. London A 209, 291 (1951)).
I summarized Dirac's derivation elsewhere as follows.
Dirac considers the following conditions of stationary action for the free electromagnetic field Lagrangian subject to the constraint $A_\mu A^\mu=k^2$: \begin{equation}\label{eq:pr1} \Box A_\mu-A^\nu_{,\nu\mu}=\lambda A_\mu, \end{equation} where $A^\mu$ is the potential of the electromagnetic field, and $\lambda$ is a Lagrange multiplier. The constraint represents a nonlinear gauge condition. One can assume that the conserved current in the right-hand side of the equation is created by particles of mass $m$, charge $e$, and momentum (not generalized momentum!) $p^\mu=\zeta A^\mu$, where $\zeta$ is a constant. If these particles move in accordance with the Lorentz equations \begin{equation}\label{eq:pr2} \frac{dp^\mu}{d\tau}=\frac{e}{m}F^{\mu\nu}p_\nu, \end{equation} where $F^{\mu\nu}=A^{\nu,\mu}-A^{\mu,\nu}$ is the electromagnetic field, and $\tau$ is the proper time of the particle ($(d\tau)^2=dx^\mu dx_\mu$), then \begin{equation}\label{eq:pr3} \frac{dp^\mu}{d\tau}=p^{\mu,\nu}\frac{dx_\nu}{d\tau}=\frac{1}{m}p_\nu p^{\mu,\nu}=\frac{\zeta^2}{m}A_\nu A^{\mu,\nu}. \end{equation} Due to the constraint, $A_\nu A^{\nu,\mu}=0$, so \begin{equation}\label{eq:pr4} A_\nu A^{\mu,\nu}=-A_\nu F^{\mu\nu}=-\frac{1}{\zeta}F^{\mu\nu}p_\nu. \end{equation} Therefore, the last three equations are consistent if $\zeta=-e$, and then $p_\mu p^\mu=m^2$ implies $k^2=\frac{m^2}{e^2}$ (so far the discussion is limited to the case $-e A^0=p^0>0$).
Thus, the first equation with the gauge condition \begin{equation}\label{eq:pr5} A_\mu A^\mu=\frac{m^2}{e^2} \end{equation} describes both independent dynamics of electromagnetic field and consistent motion of charged particles in accordance with the Lorentz equations. The words "independent dynamics" mean the following: if values of the spatial components $A^i$ of the potential ($i=1,2,3$) and their first derivatives with respect to $x^0$, $\dot{A}^i$, are known in the entire space at some moment in time ($x^0=const$), then $A^0$, $\dot{A}^0$ may be eliminated using the gauge condition, $\lambda$ may be eliminated using the first equation for $\mu=0$ (the equation does not contain second derivatives with respect to $x^0$ for $\mu=0$), and the second derivatives with respect to $x^0$, $\ddot{A}^i$, may be determined from the first equation for $\mu=1,2,3$.
However, the above is about classical electrodynamics. What about quantum theory? It turns out that modified Maxwell equations can be equivalent to the Klein-Gordon-Maxwell electrodynamics or (with some caveats) to the Dirac-Maxwell electrodynamics (see my article Eur. Phys. J. C (2013) 73:2371 at https://link.springer.com/content/pdf/10.1140/epjc/s10052-013-2371-4 ).