TITLE: Do there exist two primes $p<q$ such that $p^n-1\mid q^n-1$ for infinitely many $n$?
QUESTION [20 upvotes]: We can prove that there is no integer $n>1$ such that $2^n-1\mid 3^n-1$. This leads to the following question:

Is it true that for every pair of primes $p<q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$? Are there two primes $p<q$ and an integer $n>p+q$ such that $p^n-1\mid q^n-1$?Is it true that if $n>6$ then $2^n-1\nmid 5^n-1$?

Edit: Here are some examples:
\begin{array}{ll}
2^{36}-1\mid 41^{36}-1,
&3^{12}-1\mid 97^{12}-1,\\
5^{6}-1\mid 37^{6}-1,
&7^{4}-1\mid 151^{4}-1.
\end{array}
Now I prove there is no integer $n>1$ such that $2^n-1\mid 3^n-1.$
Proof: Denote $A=2^n-1$ and $B=3^n-1$. If $n$ is even then $3$ divides $A$ but not $B$, a contradiction.
If $n$ is odd then $A\equiv -5\pmod {12}$. Since every prime greater than $3$ is $\equiv \pm1,\pm5 \pmod {12}$, some prime factor $p$ of $A$ must be congruent to $\pm5 \pmod {12}$. As $p \mid B$, we have $3^{n+1}\equiv 3 \pmod p$. Since $n+1$ is even, we get $(\frac{3}{p})=1$, a contradiction again.

REPLY [4 votes]: Question : Is it true that for every pair of primes $p<q$ there are only finitely many integers $n$ such that $p^n-1\mid q^n-1$ ?
Answer. yes.

I will not provide a proof for this theorem here, but I will point to some references for the proof. Actually a generalization of your claim is true: for ever pair of integers $a<b$ there are only finitely many integers $n$ such that $a^n-1$ divides $b^n-1$ unless there $b=a^k$ for some integer $k$. In this document you will find the following claim proved (page 427):
Claim : If $(a^n-1)/(b^n-1)\in \mathbb Z $ for infinitely many $n\in \mathbb N$ then $b$ is a power of $a$.
The proof uses only some analysis and the Subspace theorem due to Wolfgang M. Schmidt (1972) which has many applications in number theory.
Remark although  this gives an answer to your first question there may exist an elementary proof for this because the hypothesis $p,q$ are primes is very strong.