When does $2^n-1$ divide $3^n-1$? 
Is it possible for some integer $n>1$ that $2^n-1\mid 3^n-1$ ?

I have tried many things, but nothing worked.
 A: I was looking for this as well, and eventually figured it out myself. So here's my solution for future reference. The short answer is, $2^n - 1$ never divides $3^n - 1$. Here's the proof, making use of the Jacobi symbol.
Assume $2^n - 1 \mid 3^n - 1$. If $n = 2k$ is even, then $2^n - 1 = 4^k - 1 \equiv 0 \bmod 3$. Consequently, $3$ must also divide $3^n - 1$, which is a contradiction. At the very least, we can already assume $n = 2k + 1$ is odd. Next, since $3^n \equiv 1 \bmod 2^n - 1$, from the properties of the Jacobi-symbol it follows that 
\begin{equation}
1 = (\frac{1}{2^n - 1}) = (\frac{3^n}{2^n - 1}) = (\frac{3^{2k}}{2^n - 1}) \cdot (\frac{3}{2^n - 1}) = (\frac{3}{2^n - 1})
\end{equation}
However, using Jacobi's law of reciprocity we also know
\begin{equation}
(\frac{2^n - 1}{3}) = (\frac{3}{2^n - 1}) \cdot (\frac{2^n - 1}{3}) = (-1)^{\frac{3 - 1}{2}\frac{2^n - 2}{2}} = (-1)^{2^{n - 1} - 1} = -1
\end{equation}
The only quadratic non-residue $\bmod 3$ is $2$, therefore $2^n - 1 \equiv 2 \bmod 3$ or alternatively $2^n \equiv 0 \bmod 3$. Since this implies $3$ divides $2^n$, we again arrive at a contradiction.
A: When $2^n-1$ is a Mersenne prime,this can be resolved ( although this isn't very helpful, because we only know of 49 Mersenne primes and we don't know if they are finitely many.However, it sure is nice to know that $ 2^{74,207,281} − 1$ does not divide  $3^{74,207,281} − 1$).
Let $q = 2^p-1$ be prime, therefore $F_q$ is a field. We know that polynomials of degree k must have at most k solutions in a field.Applying this to $x^p-1$, which has the solution 2 mod q, we see that this must have at most p solutions.But the set $A=(1,2,...,2^{p-1})$ obviously consists of different solutions, therefore it is the complete solution set. Since $q|3^p-1$ , we see that 3 is a solution, therefore $3 \in A $, but all the elements of the set $A-3$ have modulus less than q (obviously) and are different from 0, so no such solution may exist.
When n is a prime, but $2^n-1$ is not necessarily a Mersenne prime, we can employ the same reasoning for a prime divisor $q$ of $2^n-1$ :3 must be congruent to some power of 2 modulo q. Therefore q divides a number of the form $2^i-3$.I don't know what the prime divisors of the sequence $2^i-3$ are, but a very weak corollary is this : either $3$ or $6$ is a quadratic residue mod q, therefore, by toying with quadratic reciprocity a bit, we get this  : $q \equiv \pm 1, \pm 5, \pm 13\pmod{24}$.So when n is prime, the prime divisors of $2^n-1$ must be of this specific form (note that this is a very weak corollary).
