How can an inverted anharmonic potential $V(x)=-x^4$ have discrete bound states?
If you have a reflecting boundary condition at infinity ($\psi_{\infty}=0$), then the particle crosses the distance between this potential and infinity for a finite time (infinite velocity for this potential is reached quickly in a non relativistic case). A finite-time periodic motion is quantized in QM, like in Bohr-Sommerfeld quantization of a quasi-periodical motion.
More generally, Carl Bender et al. are considering $PT$-symmetric Hamiltonians of the form
$$ H~=~ p^2 + x^2 (ix)^{\varepsilon}, \qquad \varepsilon\in\mathbb{R} ,$$
cf. e.g. Refs. 1-3. The Hamiltonian $H$ is not self-adjoint in the usual sense, but self-adjoint in a $PT$-symmetric sense. OP's case corresponds to $\varepsilon=2$. The trick is to analytically continue the wave function $\psi$ with real 1D position $x\in\mathbb{R}$ into the complex position plane $x\in\mathbb{C}$, and prescribe appropriate boundary behaviour in the complex position plane.
See e.g. Refs. 1-3 and references therein for further details and applications. Note that Refs. 1-3 mainly discuss the point spectrum of the operator $H$.
References:
C.M. Bender, D.C. Brody, and H.F. Jones, Must a Hamiltonian be Hermitian?, arXiv:hep-th/0303005.
C.M. Bender, Introduction to $PT$-Symmetric Quantum Theory, arXiv:quant-ph/0501052.
C.M. Bender, D.W. Hook, and S.P. Klevansky, Negative-energy $PT$-symmetric Hamiltonians, arXiv:1203.6590.