When does $(a^p-1)/(a-1)$ have a $p$th power factor?

Question

Today I noted that $(18^3-1)/(18-1) = 343 = 7^3$, and that there are no other solutions to the equation $(a^3-1)/(a-1) = b^3$ with $b \le 100000$. There are, however, many solutions to the equation $$ \frac{a^p-1}{a-1} = cb^p \qquad(\star) $$ for some integer $c \ge 1$, when $p=3$.

When $p=5$, there are far fewer solutions to $(\star)$, two of which are $a=37107$ and $a=46709$.

My question is: Under what conditions does $(\star)$ have any solutions? This is clearly related to problems like Fermat's Last Theorem and Catalan's Conjecture, but is not nearly as restrictive, e.g., I'm not requiring that $a-1$ also be a $p$th power.

Answer 1

This should be a comment but is too long for the comment-functionality

The following 3 examples are from my small treatize on that "fermatquotient(Wikipedia)"-problem, done in Pari/GP, where the "%" sign means the "mod"-operation, and should (and does) result in "zero" in this cases:

   a=324;         
   (a^3-1)/(a-1) % 7^3            \\ == 0

   a= 571634088997719       
   (a^11-1)/(a-1) % 23^11         \\ == 0

   a=4224889596704250828327920681323525885423422525151934668362656 \ 
      5404363754705104377419988318310806044652742101150754590906889 \ 
      1170830410597973985409981849727159498907290190733704144470440 \ 
      4036891977239754652500794951539355221519064613575802650426875 \ 
      6338497009455353525052 
   (a^113-1)/(a-1) % 227^113     \\ == 0

There are arbitrarily many such cases but the scheme is a bit too complicated to do it here, but thats what my treatize is exactly about. Here is the "initial observation" to show how it proceeds;