Prove $\forall n\ge 2, 2^{n}-1 \nmid \ 3^{n}-1$ How can I prove for $n\ge 2$ that
$$ 2^{n}-1 \nmid \ 3^{n}-1$$
I have not been able to make any progress on this whatsoever. I would appreciate if someone can give me a hint!
 A: If $n$ is even, consider modulo $3$.
Assume $n$ is odd, and assume for contradiction that the division holds. If $p=4k+3$ is a prime divisor of $2^n-1$, then it is also a prime divisor of $3^n-1$. Using this and quadratic reciprocity, can you show that $p=3l+2$ for some integer $l$? Similarly, if $p=4k+1$, show that $p=3l+1$ for some $l$.
This means the prime divisors $\equiv 3\pmod 4$ of $2^n-1$ are exactly those $\equiv 2\pmod 3$. Since $2^n-1\equiv 3\pmod 4$, there should be an odd number of them (counting multiplicities), which means they multiply to $2\pmod 3$. This means $2^n-1\equiv 2\pmod 3$ as well, a contradiction.
A: (This is more a comment than an answer) 
I'd like to relate the problem to a well known (but unsolved) conjecture, from where the question would immediately be answered - from a consequence of the famous Waring-problem (see for instance mathworld, "power fractional parts").
 We rearrange the given expression and adapt it to the similar expression which occurs in the Waring-problem:
$$\Large \begin{array}{} {3^n-1 \over 2^n-1} &= {3^n \over 2^n-1}-{1 \over 2^n-1} \\
 &= {3^n \over 2^n}+{3^n \over 4^n}+{3^n \over 4^n}\left({1 \over 2^n}+{1 \over 4^n}...\right)-1\left({1 \over 2^n}+{1 \over 4^n}...\right) \\
&={3^n \over 2^n}+{3^n \over 4^n} - \left(1-{3^n \over 4^n}\right){1 \over 2^n-1} \\
{3^n-1 \over 2^n-1} & \lt {3^n \over 2^n}+{3^n \over 4^n} \end{array}
$$
The comparision on the rhs of the "$\lt$" reflects a detail of the Waring-problem: there we have the conjecture (for $n \gt 6$), that 
$$ \Large \begin{array}{}
  & \Big \lfloor {3^n \over 2^n} \Big \rfloor & \lt &{3^n \over 2^n}-{3^n \over 4^n} & \qquad \small ( \text{ and of course } \lt{3^n \over 2^n} )\\
\small \text{and} &&& {3^n \over 2^n}+{3^n \over 4^n} &\lt \Big \lceil {3^n \over 2^n} \Big \rceil \end{array}$$
So, if the Waring-conjecture holds, we have that $  {3^n-1 \over 2^n-1} $ is smaller than the next greater integer above   $  {3^n \over 2^n} $ so we get immediately that $2^n-1 $ does never evenly divide $3^n-1$ when $n \gt 6$  .       
I find this a nice and challenging conjecture.
Unfortunately the nice proof of @pi66 does not in reverse help for the truth of the Waring-conjecture...
