Well, I'm also a bit unsure, how to treat the hydrogen atom for first-learners. At the moment I do first the angular momentum eigenvalue problem (i.e., finding the common eigenvectors of ##\hat{\vec{J}}^2## and ##\hat{J}_3## from the commutator relations alone, leading to ##j \in \{0,1/2,1,\ldots \}## and ##m_J \in \{-J,-J+1,\ldots,J-1,J \}## with eigenvalues ##j(j+1) \hbar^2## and ##m_J \hbar##.
Then I argue that for the orbital angular momentum ##\vec{L}=\hat{\vec{x}} \times \vec{p}## there are only integer solutions, because of the meaning of the rotations ##\exp(-\mathrm{i} \hat{\vec{L}} \cdot \vec{\phi})## on the position eigenfunctions. Alternatively one can also argue with the 2D symmetric harmonic oscillator, where you can directly proove that ##\hat{L}_3## has only eigenvalues ##m \hbar## with ##m \in \mathbb{Z}##.
Finally, I write down the time-independent Schrödinger equation and solve for the radial wave function in
the ansatz ##\psi_{n\ell m}=R_{n \ell}(r) \text{Y}_{\ell m}(\vartheta,\varphi)##. This is, however, a quite lenghty calculation, starting with a discussion of the boundary conditions for ##r \rightarrow 0## and ##r \rightarrow \infty## and then the result that the energy eigenvalues are ##E_n=-1 \text{Ry}/n^2## with ##n \in \mathbb{N}##. Then you have the somewhat astonishing additional degeneracy of the energy eigenvalues, i.e., that ##E_n## and ##R_{n \ell}## in fact only depend on ##n## and not on a combination of ##n## and ##\ell##, which is specific for the Coulomb potential. Nevertheless it's a nice non-trivial example for the solution of a Sturm-Liouville boundary-value problem of a differential equation and the definition of the corresponding CONS of eigenfunctions.
This is well explained when using the Runge-Lenz vector which is known from classical mechanics as an application of Noether's theorem. The only trouble is that of course you need anglar-momentum addition, i.e., the Clebsch Gordan coefficients, because indeed the SO(4) symmetry (for the bound states) is solved by using ##\mathrm{so}(4) \equiv \mathrm{su}(2) \oplus \mathrm{su}(2)##.
Another advantage of this analysis is that you also get the scattering states for ##E=0## (with the symmetry group ISO(3)## and ##E>0## (with the symmetry group SO(1,3)).
