Does the equation $x^m-1=y^n+y^{n-1}+...+1$ have only finitely many solutions in prime powers $x,y$ with $y>2$ and integers $m,n>1$?

This question arose in the study a group theory problem, where I'm examining a "natural" subgroup $G$ of $S_d$ where $d:=x^m-1$. I can show that $G$ is primitive and contains a $d$-cycle, so if $d$ is composite then known results imply that $G$ must be either $S_d$ or $A_d$ or a group between $\text{PGL}_{n+1}(y)$ and its automorphism group, for some $(n,y)$ satisfying the above equation. I'm hoping to show that the last case almost never happens. The best I can do so far is the simple observation that if $m=2$ then the set of prime $x$'s for which the equation has a solution is a density-zero subset of the primes. But I'm hoping that a much stronger result is possible.

Does the equation $x^m-1=y^n+y^{n-1}+...+1$ have only finitely many solutions $(x,y,m,n)$ where $x,y$ are prime powers with $y>2$ and $m,n$ are integers with $m,n>1$?

This question arose in the study a group theory problem, where I'm examining a "natural" subgroup $G$ of $S_d$ where $d:=x^m-1$. I can show that $G$ is primitive and contains a $d$-cycle, so if $d$ is composite then known results imply that $G$ must be either $S_d$ or $A_d$ or a group between $\text{PGL}_{n+1}(y)$ and its automorphism group, for some $(n,y)$ satisfying the above equation. I'm hoping to show that the last case almost never happens. The best I can do so far is the simple observation that if $m=2$ then the set of prime $x$'s for which the equation has a solution is a density-zero subset of the primes. But I'm hoping that a much stronger result is possible.