As Mateusz remarked, $V$ must be assumed to have a meromorphic extension to $\mathbb C$, and $V = \lambda + \psi''/\psi$. Now if $\psi$ has a pole or zero at $z=p$, so that $\psi(z) \sim c (z-p)^d$ as $z \to p$ (with $d \ne 0$), then
$V(z) \sim d (d-1) (z-p)^{-2}$. Thus the only poles that $V$ may have are
poles of order $2$ with leading coefficient $d (d-1)$ for an integer $d$, or (corresponding to simple zeros of $\psi$) poles of order $1$.

EDIT: When $V$ has a pole of order $1$ or $2$ at $z=p$, the differential equation has a regular singular point there.

In the case of order $1$ the indicial equation is $r^2 - r = 0$, so there is a solution with a simple zero at $z=p$ (corresponding to indicial root $1$); the other solutions (not constant multiples of this one) will have a logarithmic branch point at $z=p$.

In the case of order $2$ with leading coefficient $d(d-1)$, the indicial equation is $r^2 - r = d^2 - d$, so if $d$ is an integer $\ge 2$ there is a solution with a pole of order $d$ at $z=p$ (corresponding to indicial root $d$); the other solutions (not constant multiples of this one) may or may not have a logarithmic branch point.

If there is more than one pole of $V$, there may be compatibility conditions needed to ensure that a solution that works at one pole corresponds to solutions that work at the other poles, rather than those that have branch points.