Does helium have any accidental degeneracies? Does helium have any accidental degeneracies, i.e. are there reducible eigenspaces of $$-\Delta_1 - \Delta_2 - \frac{2}{r_1} - \frac{2}{r_2} + \frac{1}{r_{12}}~?$$ 
 A: No, it doesn't. Nor does hydrogen, for that matter. Accidental degeneracies essentially never occur in natural systems with a finite number of particles.
Essentially all nontrivial degeneracies in non-contrived quantum systems are the result of symmetries; i.e. if two energy eigenstates $|\psi\rangle$ and $|\phi\rangle$ are degenerate, then they are related by $|\phi\rangle = U |\psi\rangle$ for some unitary representation $U$ of a symmetry operator that commutes with the Hamiltonian. While this is somewhat a question of terminology, I would say that such a degeneracy is "symmetry-induced" rather than "accidental".
An example of an accidental symmetry would be a particle in a 2D box with aspect ratio $L_y/L_x = \sqrt{3/5}$. The $(n_x = 3, n_y = 1)$ and $(n_x = n_y = 2)$ energy eigenstates both happen to have energy
$E_{3,1} = E_{2,2} = 16 \pi^2 \hbar^2/(3 L_x^2)$, so they are accidentally degenerate, but there is no global symmetry of the Hamiltonian that maps either state to the other. Like all accidental degeneracies, this one is fine-tuned and immediately vanishes if we slightly perturb the Hamiltonian, and it has no connection to any conserved quantities. Such arbitrarily fine-tuned conditions rarely occur in nature.
A: Well, at least helium ion He$^+$ has the same accidental degeneracy as the hydrogen atom (although the degeneracy is only exact in the nonrelativistic limit for both of them).
EDIT (September 21, 2022): The above can also be adapted to the neutral He atom as follows. Let us consider the states of the neutral He where one of the electrons is in a highly excited state (still bound, but close to a free state). As this electron can be pretty far from the nucleus, it will not influence the state of the other electron much, so such states of the neutral He will be somewhat similar to the states of the hydrogen atom:

In general, at sufficiently high principal quantum numbers, an excited
electron - ionic core system will have the general character of a
hydrogenic system and the energy levels will follow the Rydberg
formula.

Furthermore, let us consider two almost degenerate states of the He$^+$ ion and add to each of them an electron in a "high" Rydberg state. As the spectrum of "high" Rydberg states is pretty dense, one can probably choose the states of the Rydberg electrons in such a way that the difference in their energies would compensate the inaccuracy of the degeneracy. Let me also add that spectral lines (and all energy levels except for the ground level) have natural width, which can simplify finding degeneracy. So it looks like one can find states of the neutral He atom that are closer to degeneracy than any states of the hydrogen atom.
