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?

Note for this type of question is always useful to ask what how many times does the denominator easily go into the numerator, here we have $2^a/2^b = 2^{a-b}$, thus $$\frac{2^a + 1}{2^b -1} = \frac{2^a + 1 - 2^{a-b} (2^b - 1)}{2^b - 1} + 2^{a-b} = \frac{2^{a-b} + 1}{2^b - 1} + 2^{a-b}$$
Thus if $a > b$, then 
$$\frac{2^a + 1}{2^b -1} = \frac{2^{a-b} + 1}{2^b - 1} + \mbox{some integer}$$
And this jumps out to me as something that can be repeated.
It is easy to show that similarly for any $x$ with $x > b$, $$\frac{2^{x} + 1}{2^b - 1} = \frac{2^{x-b} + 1}{2^b - 1} + \mbox{some integer}$$
Thus we can repeat this process, "subtracting" $b$ from the power in the numerator and get
$$\frac{2^a + 1}{2^b -1} = \frac{2^{r} + 1}{2^b - 1} + \mbox{some integer}$$
where $r < b$.
The rest is just inequalities to show that this can never be equal to an integer.
Now as $r < b$, thus $r \leq b-1$, thus $2^r \leq 2^{b-1}$.
Also $2^{b-1} < 2^b - 2$, thus $2^r + 1 < 2^b + 1$, thus $$0< \frac{2^{r} + 1}{2^b - 1} < 1$$
Thus $$\frac{2^a + 1}{2^b -1} = \mbox{non integer} + \mbox{some integer}$$
