<s>[INST] 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.
 [/INST] 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.
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