if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then does $a | 2^m-1$ or $a| 2^m+1$ 
Determine whether for any integer $a$, if $k > 1$ and $m\ge 1$ are such that $a^k = (2^m-1)(2^m+1)$, then $a | 2^m-1$ or $a| 2^m+1$.

Suppose $a\not |2^m-1$ and $a\not| 2^m+1$. Then there exist primes $p_1$ and $p_2$ so that $p_1 | 2^m-1, p_1 \not | 2^m+1, p_2 | 2^m+1, p_2 \not | 2^m-1.$ Since $2^m+1$ and $2^m-1$ are coprime, so are $p_1$ and $p_2$. By Bezout's identity, there exist integers $s,t$ so that $sp_1 + tp_2 = 1$. Since $p_1 | 2^m - 1$ and $p_2 | 2^m + 1$ there exist integers $c,d$ with $cp_1 + 1 = 2^m, dp_2 - 1= 2^m$ and so $dp_2 - cp_1 = 2$. Since $k > 1, p_1^2 | 2^m-1$ and $p_2^2 | 2^m+1$ and similarly there exist integers $c,d$ with $cp_1^2 + 1 = 2^m, dp_2^2 - 1= 2^m$ and so $dp_2^2 - cp_1^2 = 2$. In fact, both $2^m+1$ and $2^m-1$ must have prime factorizations where the exponent of each prime is at least $2$; if one prime factor of either has an exponent of one, then it is not possible that $a^k = (2^m-1)(2^m+1)$ because all prime powers in the prime factorization of $a^k$ have exponent at least $2$.
But I'm not sure how to get a contradiction from this. Maybe there's a counterexample?
I also know the following theorem: If $a$ and $b$ are positive integers, then there exists a number $g$ such that every multiple of $\gcd(a,b)$ greater than $g$ may be written in the form $ra+sb$ where $r$ and $s$ are nonnegative integers, but I don't think it's necessary for this problem.

Let $a$, $k$ and $m$ be positive integers with $k\geq2$ and
$$a^k=(2^m-1)(2^m+1).$$
The two factors on the right hand side are coprime, because they are both odd and their difference is $2$. Then by unique factorization, both are $k$-th powers. That is, there exist positive integers $b$ and $c$ such that
$$2^m-1=b^k\qquad\text{ and }\qquad 2^m+1=c^k.$$
Subtracting the former from the latter shows that
$$c^k-b^k=2,$$
but of course there are no two positive powers that differ by $2$.
So the answer is vacuously yes, because there are no such integers.
