I assume $\Omega$ is a bounded open subset of $\mathbb{R}^n$ and, say, $V\in L^\infty(\Omega)$. What you say is correct, but of course you need to assume that the eigenvalues are also consecutive (otherwise e.g. $V$ itself could be another eigenvalue in between). The comparison principle you are seeking comes from the Courant–Fischer–Weyl variational characterization of the eigenvalues of $-\Delta + V,$ by which the $k$-th eigenvalue $\lambda_k(-\Delta + V\\ )$ in the increasing order is expressed as a certain min-max of the Rayleigh quotient, which is monotone wrto $V:$ $$\frac{\int_\Omega \left(|\nabla u|^2+V(x)u^2 \right)dx }{\int_\Omega u^2 dx }$$ (for a precise statement see e.g. Courant-Hilbert, or Reed-Simon, or Gilbarg-Trudinger, &c).
So if we denote $\lambda_k:=\lambda_k(-\Delta)$ and assume $-\lambda_{k+1} < V(x) < -\lambda_{k}\\ ,$ it follows by the monotonicity $$\lambda_k(-\Delta+V)<\lambda_k(-\Delta - \lambda_k)=0 =\lambda_{k+1}(-\Delta - \lambda_{k+1})< \lambda_{k+1}(-\Delta +V),$$ so that $0$ is not an eigenvalue of $-\Delta +V.$