Proving $\{0^{2^n}\}$ is not regular using pumping lemma
If you choose $k = 2$ then, writing $|y| = m$, we get
$$ xy^2z = 0^{2^n+m}. $$
Since $1 \leq m \leq n$, we have
$$ 2^n &lt; 2^n+m \leq 2^n+n. $$
Cantor's theorem shows that $n &lt; 2^n$, and in particular,
$$ 2^n &lt; 2^n+m &lt; 2^{n+1}. $$
Therefore $xy^2z \notin L$.

Let me take this opportunity to mention another misconception. There are some non-regular languages which satisfy the pumping lemma. Therefore the fact that you couldn't prove that a certain language is non-regular using the pumping lemma does not imply that the language is regular.