Consider the equation $P(x)=y^d$ where $d \geq 2$ is an integer and $P$ can be written $P(x)=c(x-r_1)(x-r_2)\ldots (x-r_t)$ where $c$ and all the $r_i$ are integers not all equal (some of them can be equal however). I say that the equation is regular if all the integer solutions are trivial (in the sense that $y=0$), and singular otherwise.
Thus, the Erdos-Selfridge theorem states that the equation is regular when $c=1$ and the $r_i$ form an arithmetic progression with common difference one. Is it true that for a fixed $P$, there are only finitely many $d$ for which $P(x)=y^d$ is singular ?
A theorem of Schinzel and Tijdeman from 1976 implies that if $P(x)$ is a polynomial with integer coefficients and at least two distinct (complex) roots, then there exists an effectively computable constant $d_0$ such that the equation $P(x)=y^d$ has no solutions with $|y|>1$ and $d > d_0$. This answers your question in the affirmative provided one rules out those $P(x)$ for which the equation $P(x)=\pm 1$ is solvable, such as $P(x)=x(x-2)$, which are clearly (in your notation) singular for all $d$.
The result of Schinzel and Tijdeman relies upon lower bounds for linear forms in logarithms. A proof can be found in the book of Shorey and Tijdeman.