For the first question, I leave it up to the experts in point spectra and stuff like that (even though I think that you only need $V(x)\to\infty$ for $|x|\to\infty$).
Regarding your second question: spectra with $E_n = n^c$. I present below the reason why $c>2$ should be impossible. In fact you can easily find a potential where $E_n \sim n^c$ for $n\to\infty$ by using the correspondence principle:
The differences of energies scale like $\Delta E_n \sim n^{c-1}$ which should correspond to $h/T$ where $T$ is the period of the classical motion in the potential. Given that
$$ T(E) = \int_{V\leq E} \frac{dx}{\sqrt{2(E- V)/m}} $$
we find that with $V(x) \sim x^{\alpha}$, we have $T(E) \sim R^{(\alpha-2)/\alpha} \sim E^{(2-\alpha)/2\alpha}$ with $R$ the turing point such that $V(R) = E$.
So in order to have $E_n = n^c$, we need to have
$\Delta E_n \sim n^{c-1} \sim E^{(c-1)/c}$
to match $T^{-1} \sim E^{(\alpha-2)/2\alpha}$ which leads to
$$\frac{c-1}{c} = \frac{\alpha-2}{2\alpha}$$
or
$$\alpha= \frac{2c}{2-c}.$$
For $c\to 2^-$, we have $\alpha$ approaching $\infty$. So for the steepest possible potential, we only get $c=2$.
Note that:
- for $c=1$, we get $\alpha=2$ (harmonic oscillator)
- for $c=-2$, we get $\alpha=-1$ (H-Atom)
By the way, looking at the inverse relation:
$$c = \frac{2\alpha}{2+\alpha}$$
we see that there is also something happening at $\alpha=-2$: this is the point at which Heisenbergs uncertainty principle is not enough to hinder the particle from falling into the centre.
