Prove that $\frac{2^a+1}{2^b-1}$ is not an integer 
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.

This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we prove the statement from this or is there an easier way?

Assume that $2^b-1$ is a divisor of $2^a+1$ and write $a$ as $kb+r$ with $0\leq r<b$. Then:
$$2^a+1\equiv (2^b)^k\cdot 2^r+1 \equiv 2^r+1 \pmod{(2^b-1)} $$
but since $r<b$ and $b>2$, $2^r+1$ is too small to be $\equiv 0\pmod{2^b-1}$.
