Are there non-constant potentials that result in eigenstates of the Hamiltonian that are all plane waves?

Plane waves are eigenstates of the momentum operator. Another observable can have the same eigenstates only if it commutes with the momentum operator (see this answer for more details).

Now, the momentum operator generates translations, so the Hamiltonian commutes with it if it is invariant under translations. The kinetic part of the Hamiltonian is invariant (i.e. it commutes with the momentum operator). The potential in general not. The only potential that commutes with the momentum operator is the flat potential.

In summary: if $H$ has to have the same eigenstates as $p$, then you must have that $[H,p]=0$. This is equivalent to $[p^2/2m+V,p]=[V,p]=0,$ which tells you that $V=mathrm{const}$.

No, there are no other potentials with the same eigenstates. (set $hbar = m = 1$ for clarity)

Suppose there is another Hamiltonian $hat{H}'$, such that the eigenstates of $hat{H}$ and $hat{H'}$ are the same, the complete, orthogonal set $| psi rangle$ (obtained via the spicy spectral theorem). Then, we know that $[H, H'] = 0$, since $hat{H} hat{H}' |psi rangle = E' hat{H} |psi rangle = E' E |psi rangle$. Neat!

Then we can do this: (Note here that $V$ is an analytic function, and by "$V(hat{x})$", I mean $sum_{k=0}^infty frac{1}{k!} frac{partial V(0)}{partial x} hat{x}^k$ $$ 0=[hat{H},hat{H}'] = [frac{1}{2}hat{p}^2 + V, frac{1}{2}hat{p}^2 + V'] = [frac{1}{2}hat{p}^2, V'] + [V, frac{1}{2}hat{p}^2]\ Rightarrow [hat{p}^2, V(hat{x})] = [hat{p}^2, V'(hat{x})] $$ That's pretty big, since that translates (after a bit of algebra, involving expanding V in a Taylor series and using $[x, p] = i$ ) to $$ -2i frac{partial V}{partial x}hat{p} - frac{partial^2 V}{partial x^2} = -2i frac{partial V'}{partial x}hat{p} - frac{partial^2 V'}{partial x^2} $$

By substituting in e.g. momentum eigenstates, this condition begins to look like $$partial_xV'(x) = partial_xV(x)$$ By simply integrating this here equation, we see that $V-V' = constant$.