You're confusing two things:
- The discreteness or "quantization" of available physical states, and
- the quantization of excitations in those states.
The first thing exists in classical physics and does indeed arise from boundary conditions.
The second thing is truly quantum.
It's not your fault that you're confused about this.
It's confusing because books, online videos, etc. all explain it terribly, if they bother to explain it at all.
Let's check my statement that the discreteness of physical states comes from boundary conditions and exists in classical physics.
Consider a violin string, described by a function $u(x, t)$ where $u$ is the displacement of the string.
The string is clamped on both ends (that's the boundary condition) so
$$u(x=0, t) = u(x=L, t) = 0 \, .$$
As you probably know, the displacement of the string can be expressed as a Fourier series
$$u(x, t) = \sum_{k=0}^\infty c_k(t) \sin(\pi k x / L) \, .$$
There are your discrete physical states right there: they have shape $\sin(\pi k x / L)$ and are enumerated by the integer $k$.
We've written the general state of the string $u(x, t)$ as a discrete sum over these other states, each with a time dependence $c_k(t)$.
This is exactly like expressing a general electron quantum state as
$$\left \lvert \psi(x, t) \right \rangle = \sum_{k} c_k(t) \left \lvert \psi_k(x) \right \rangle$$
where the set of states $\left \lvert \psi_k \right \rangle$ are something usually chosen to have simple time dependence, i.e. something such that $c_k(t)$ are simple.
Of course, the simplest choice of $\left \lvert \psi_k \right \rangle$ are those such that $H \left \lvert \psi_k \right \rangle = E_k \left \lvert \psi_k \right \rangle$ because then $c_k(t) = \exp(-i E_k t / \hbar)$.
We do exactly the same thing in classical physics: the Fourier series is convenient precisely because once you substitute it into the wave equation that describes the string
$$\frac{\partial^2 u}{\partial t^2} = v^2 \frac{\partial^2 u}{\partial x^2}$$
you get
\begin{align}
\sum_{k=0}^\infty \ddot{c}_k(t) \sin(\pi k x / L)
&= -v^2 \sum_{k=0}^\infty c_k(t) \left( \frac{\pi k}{L} \right)^2 \sin(\pi k x / L)\\
\ddot{c}_k(t) &= - \left( \frac{\pi v k}{L} \right)^2 c_k(t) \\
\rightarrow c_k(t) &= a_k \sin\left(\pi v k t / L \right) + b_k \cos\left(\pi v k t / L \right)
\end{align}
i.e. simple sinusoidal time dependence.
This is exactly the same procedure you use to solve a quantum particle in a well or the hydrogen atom (or any physical system described by linear time invariant differential equations).
These nice shapes of the string $\sin(\pi k x / L)$ which have simple time dependence are called "normal modes".
Now here's where classical and quantum mechanics differ:
In classical mechanics, the numbers $a_k$ and $b_k$, i.e. the things that say how excited each normal mode is, are continuous numbers, whereas in quantum mechanics their squares $n = a^\dagger a$ have a discrete spectrum.
That's the quantum part of quantum mechanics.
Now I invite you to read this other question which explains why you should not think of each excitation of a field (like an electron) as being a particle with its own identity, but should rather think of electrons as identity-less units of excitation like our classical $a_k$ and $b_k$.
