classical mechanics - Relation between (super)integrability and closed orbits

Inspired by this recent question, I would like to understand from a more general and mathematical perspective why closed orbits are only found for the Kepler ($V(r) \sim 1/r$) or harmonic ($V(r) \sim r^2$) potential problems, as follows from Bertrand's theorem.

There are two aspects that make these problems special, which I suspect may be related to the closed-orbit property. First, both problems are superintegrable. This property sits intuitively well with the idea that phase-space orbits should close "as quickly as possible", thus implying that real-space orbits close after a single revolution. Second, each problem possesses an additional "unexpected" conserved quantity, due to a larger symmetry of the problem than the obvious $O(3)$. For the Kepler problem, this is the Runge-Lenz vector, related to the $O(4)$ symmetry of the Hamiltonian. Meanwhile, the harmonic oscillator Hamiltonian conserves the Fradkin tensor:

$$F_{ij} = \frac{p_i p_j}{m\omega^2} + m\omega^2 q_i q_j,$$ which is related to an $SU(3)$ symmetry. In fact these symmetries and corresponding conserved quantities exist for any central field problem (D. M. Fradkin, Prog. Theor. Phys. 37 (1967), p.798). However the conserved quantities only take a "nice" form for the Kepler and harmonic problems, which also allows the corresponding quantum problems to be diagonalised exactly by symmetry arguments alone.

These considerations motivate the following question:

What specific physical/mathematical feature(s) do these two problems share that gives them the property of closed orbits? Does this feature bear relevance to the quantum counterpart?