Divisibility of integers Let $n > 1$ be an integer. Then $2^n - 1\nmid 3^n - 1$. I don't know how to prove it. Can anybody help me, please?
In general, for a fixed positive integer $a > 1$, has $a^n - 1|(a +1)^n - 1$ any integer solutions?

The idea used below is very close to the one used by @Zander. It will not be a surprise to those who have seen my other posts that the details take longer.
If $n$ is even, then $2^n-1$ is divisible by $3$, so $2^n-1$ cannot divide $3^n-1$ unless $n=0$.
So let $n>1$ be odd.  Let $p$ be a prime that divides $2^n-1$.  Then since $2^n \equiv 1 \pmod{p}$, the order of $2$ modulo $p$ is odd, so $2$ is a quadratic residue of $p$.  If furthermore $2^n-1$ divides $3^n-1$, then $3^n \equiv 1 \pmod {p}$, and therefore $3$ is also a quadratic residue of $p$.
The number $2$ is a quadratic residue of the odd prime $p$ iff $p\equiv \pm 1 \pmod{8}$.  So $p$ must be of the shape $24k+1$, $24k+7$, $24k+17$, or $24k+23$.  But by Quadratic Reciprocity, $3$ is a non-residue of $p$ if $p$ is of the shape $24k+7$ or $24k+17$.  Thus it is enough to show that if $n$ is odd, then $2^n-1$ has at least one prime factor of the shape $24k+7$ or $24k+17$. 
To do this, we show that not all primes in the prime factorization of $2^n-1$ can be of shapes $24k+1$ and/or $24k+23$.  Suppose to the contrary that they all are.  We will obtain a contradiction. 
Note that $24k+1$ is congruent to $1$ modulo both $3$ and $8$, while $24k+23$ is congruent to $-1$ modulo both $3$ and $8$.
Since $2^n-1$ has shape $8s-1$, its prime factorization must have an odd number of occurrences of (not necessarily distinct) primes of the form $24k+23$.  But that implies that $2^n-1\equiv -1 \pmod 3$, which is not the case when $n$ is odd. 
