I am however confused about how...
$$ [\hat{x}, \hat{p}] = \imath\hbar \quad\Rightarrow\quad \Delta x\Delta p \geq \frac{\hbar}{2}.$$
can be used to explain quantization of energies as claimed in SE posts like.
A simple example where the differential equation's boundary conditions can be hidden under the rug of algebraic formalism is the quantum simple harmonic oscillator. (But at the end of the day, as we will see, you still need the boundary conditions!)
... where a problem in quantum mechanics is solved using the Schrödinger equation
For the simple harmonic oscillator, you solve the Schrodinger equation:
$$
-\frac{1}{2}\frac{d^2 \psi}{dx^2} + \frac{1}{2}x^2\psi = E\psi\;,
$$
subject to the boundary conditions that $\psi(x)\to 0$ as $x\to \pm \infty$. (For ease of writing I set $\hbar=m=k=1$.)
After solving the differential equation, you end up with quantized energy levels.
...and the quantization of the resulting energies can clearly be linked to the Heisenberg uncertainty principle (or alternatively the canonical commutation relations) without invoking any boundary conditions?
In the algebraic formalism we define
$$
\hat a = \frac{1}{\sqrt{2}}(\hat x+i\hat p)
$$
and
$$
\hat a^\dagger = \frac{1}{\sqrt{2}}(\hat x-i\hat p)
$$
and we use the relation
$$
[\hat x,\hat p]=i
$$
to see that
$$
[\hat a,\hat a^\dagger]=1
$$
and so
$$
\hat H=\hat a^\dagger \hat a + \frac{1}{2}\;.
$$
And again we end up with quantized energies.
This follows since there is a lowest energy state such that
$$
\hat a|0\rangle = 0\;\;,
$$
which has energy $1/2$.
And we can create the other energy eigenstates by acting on $|0\rangle$ with $\hat a^\dagger$ like
$$
|1\rangle = \hat a^\dagger|0\rangle
$$
which has energy $3/2$. And so on in integer (i.e., quantized) steps.
But now you might rightfully wonder: Where did the boundary conditions go in the algebraic formalism?
The answer is that you still must include boundary conditions. But, in the algebraic formalism, they are included in the definitions of operators.
For example, the domain of the operator $\hat p$ in the SHO problem is defined as
$$
\mathcal{D}(\hat p) = \left\{
\psi(x)\in L^2| \frac{d\psi}{dx}\in L^2 \text{ and }\lim_{x\to\pm\infty}\psi(x)\to 0
\right\}\;.
$$
You must make such a definition because if you do not then $\hat p$ is not hermitian. To see this, consider
$$
\langle \chi|\hat p|\psi\rangle
=-i\int_{-\infty}^\infty dx \chi^*(x)\frac{d\psi}{dx}
$$
$$
=\left[\int_{-\infty}^{\infty}(-i\frac{d\chi}{dx})^*\psi(x)\right]
-i\left.\chi^*\psi\right|^{\infty}_{-\infty}\;,
$$
where the last term is not zero unless you use the boundary conditions.
So, for example, if we do not include the proper boundary conditions in the operator definition for $\hat p$ then the thing we called $\hat a^\dagger$ is not really the hermitian conjugate of the thing we called $\hat a$, which is a problem.
