Proving that language, with $|\Sigma|=1$, is irregular by Myhill–Nerode theorem
Your idea is basically correct, but your argument could be written in a clearer way. You want to say that if $i \neq j$ then $0^{2^i+1},0^{2^j+1}$ are inequivalent. You propose to consider adding the word $0^{2^i-1}$ to both, obtaining $0^{2^{i+1}},0^{2^i+2^j}$. The first word is clearly in $L$, whereas the second isn't, though an argument is required (left to you). This shows that the infinite set $\{0^{2^i+1} : i \geq 0\}$ consists of pairwise inequivalent words (modulo $L$), so $L$ is not regular.

In fact, all words $0^n$ are pairwise inequivalent modulo $L$ – see if you can prove that. (The argument is very similar.)

For languages over a unary alphabet there is a simpler characterization. Let $L = \{0^n : n \in I\}$. Then $L$ is regular if and only if $I$ is eventually periodic, which means that for some $m \geq 1$, $n + m \in I$ iff $n + 2m \in I$. The set of powers of two is not eventually periodic since it has arbitrarily large "gaps". (A formal argument is similar to your proof.)