Normally pumping lemma proofs aren't too hard; you just experiment with different values of $i$ and sort of stumble on the solution. However, this problem is really bugging me and I am spending a lot of time on it.
If I let $x=a^k$, $y=b^{2^k}$, and $z=\epsilon$, then $| y|>k$ and $xyz \in L$.
Now we have that $y=uvw$ for $u,v,w \in \{b\}^{*}$, $v \neq \epsilon$.
$xyz \in L$ if for all $i \geq 0$, $xuv^iwz \in L$.
In particular, $L$ will not be regular if $2^k+(i-1)m-m \neq 2^p$ for all $p,m \geq 1$, since $|uv^iw|$ is $2^k$.
I have been trying to find an $i$ such that if I add or subtract a constant from both sides, one side will always be even and the other always odd or vice versa, regardless of $m$; however, I have so far found this to be extremely hard.
Namely, I have been trying $i=m \pm d$ for some $d$. Then the left hand side contains a quadratic, and if I can find one with a factorization $(m\pm d)(m\pm s)$ that is always even, e.g. $(m-1)(m-2)$, then if the right hand side is odd ($2^p-1$, for instance), I will have a contradiction.
I have tried other solutions, like letting $i=2^k$ or some variation on that, but I have gotten nowhere.
Really, I would just like some guidance on what the form of $i$ should be. I would like to work the rest from there.
Let $p$ be the pumping length. Pick $k$ with $2^k>p$. Then $b^{2^k}\in L$ implies $b^{2^k}=uvw$ with $|uv|\le p$, $|v|>0$ and $uv^iw\in L$ for all $i\in\mathbb N_0$. Since all must be sequences of $b$ alone, we need that each term of the arithmetic sequence $|u|+i|v|+|w|$ is a power of $2$ for all $i\in\mathbb N$. As the distances between powers of $2$ grow arbitrarily large (larger than $|v|$) this is absurd. In fact, $|v|<p<2^k=2^{k+1}-2^k$ shows that already $uv^2w\notin L$.