The equation $x^m-1=y^n+y^{n-1}+...+1$ in prime powers $x,y$ 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.
 A: Mike Bennett's comments gives good reason to suspect that my original question is beyond what can be proved with existing techniques.  Here I record one thing that can be proved (again thanks to Mike's comment), namely that there are no solutions with $x=4$.  That is:
the equation $4^m-1=y^n+y^{n-1}+...+1$ has no solutions $(y,m,n)$ in which $y$ is a prime power with $y>2$ and $m,n$ are integers with $m,n>1$.
Suppose that $(y,m,n)$ is a solution.  First note that $y$ is odd and $n$ is even.
For, if $y$ were even then we would contradict uniqueness of base-$2$ expansion by writing the number $4^m-1$ in two different ways as the sum of distinct powers of $2$, namely as both $2^{2m-1}+2^{2m-2}+...+1$ and $y^n+y^{n-1}+...+1$.  Hence $y$ is odd, so the mod $2$ reduction of the equation $4^m-1=y^n+y^{n-1}+...+1$ implies that $n$ is even.
Next show that the polynomial $X^2-1-(Y^n+Y^{n-1}+...+1)$ is irreducible in $\mathbf{Q}[X,Y]$.  If it were reducible then it must factor as $(X-f(Y))\cdot(X+f(Y))$ for some polynomial $f(Y)$, so that $f(Y)^2=1+(Y^n+Y^{n-1}+...+1)$ is a square in $\mathbf{Q}[Y]$.  Rewrite this as $1+(Y^{n+1}-1)/(Y-1)$, or equivalently $(Y^{n+1}+Y-2)/(Y-1)$.  Then
$f(Y)$ divides both $Y^{n+1}+Y-2$ and its derivative $(n+1)Y^n+1$, so $f(Y)$ also divides
$(n+1)\cdot(Y^{n+1}+Y-2) - Y\cdot((n+1)Y^n+1)=nY-2(n+1)$, whence $\deg(f)=1$ so $n=2\deg(f)=2$.  But $Y^2+Y+2$ is not a square in $\mathbf{Q}[Y]$, contradiction.
Now we can apply Theorem 4.1(ii) of the paper "Estimates for the solutions of certain Diophantine equations by Runge's method" by A.Sankaranarayan and N.Saradha (Int.J.Number Theory 4 (2008), 475-493).  According to this result (which uses the condition that $n$ is even), if $u,v$ are integers such that $u^2-1=v^n+v^{n-1}+...+1$, then
$|u|\le 2^{3n+4}$.   But we know that $(u,v)=(2^m,y)$ is a solution, so $2^m\le 2^{3n+4}$.   On the other hand we have $y^n<4^m$, so that $y^n<4^{3n+4}$ and thus $y<4^{3+(4/n)}$.   I checked via computer that there are no solutions with $n<44$.  So assume $n\ge 44$, which implies that $y<73$.
Since $4^m=(y^n+y^{n-1}+...+y)+2$, we have $4^m\equiv 2\pmod{y}$, so the order of $2$ in $(\mathbf{Z}/y\mathbf{Z})^*$ is odd.  Since $y<73$, it follows that $y\in\{7,23,31,47,49,71\}$.  Each of these values $y$ satisfies either $y\equiv -1\pmod{4}$ or $y\equiv -1\pmod{5}$.  If $y\equiv -1\pmod{4}$ then, since $n$ is even, we obtain the contradiction
$4^m=(y^n+y^{n-1}+...+y)+2\equiv 2\pmod{4}$.  Likewise, if $y\equiv -1\pmod{5}$ then
$(y^n+y^{n-1}+...+y)+2\equiv 2\pmod{5}$, so that $4^m\equiv 2\pmod{5}$, which is impossible.
(Note: the Sankaranarayan-Saradha result is a refinement of an earlier result by P.G.Walsh; if one uses Walsh's result, then the same proof works except that one must rule out $y=73$ by a different method.  One way to do this is to note that the equation
$4^m-1\equiv 73^n+73^{n-1}+...+1\pmod{91}$ has no solutions in which $n$ is even.) 
A: This is too long for a comment so I will post it as an answer. 
More precisely, consider the equation
$$x^n-1=\frac{y^q-1}{y-1},$$
In what follows, I will assume that $q$ is prime but this condition can be weakened. Note, if a prime $r|\frac{y^q-1}{y-1}$ then by Fermat Little Theorem we have that either $r|q$ or $q|r-1$ and thus $r=qk+1.$
Suppose $y\ne 1(\mod q),$ then $\frac{y^q-1}{y-1}=\frac{y-1}{y-1}=1 (\mod q)$ and we must have $x^n-1=1(\mod q)$ or $x^n=2(\mod q).$ But $x-1|\frac{y^q-1}{y-1}$ and so  all prime divisors of $x-1$ are congruent to $1(\mod q).$ Thus, $x-1=1(\mod q)$ which implies that $2^n=2(\mod q).$ So if $2^n\ne 2(\mod q)$ our equation has no solutions.
Now, if $y=1 (\mod q),$ we have $\frac{y^q-1}{y-1}=0(\mod q)$ and we must have $x^n=1(\mod q).$ Moreover, $x^n-1=(x-1)(x^{n-1}+....+1)$ and since all prime divisors of both $x-1$ and $x^{n-1}+x^{n-2}+...+1$ are either $q$ or $r=1(\mod q),$ we must have $x-1=0,1(\mod q)$ and $x^{n-1}+....+1=0,1(\mod q).$ 
If $x=1(\mod q),$ we must have $n=1,0(\mod q).$ If $x=2(\mod q),$ we must have $2^n=1(\mod q.)$ Otherwise, there are no solutions.
For the general, if $q$ is not prime, we get that prime divisors of the right had sides come from divisors of $q$ or different arithmetic progressions with differences dividing $n.$ One may perform similar analysis in some particular cases.   
