We assume the regularity of $L$ for the sake of contradiction.
Pumping Lemma: $ \exists n \in \mathbb{N}$ for which $ \forall z \in L$ with $|z| \geq n$, $\exists u,v,w$ with $|uv| \leq n$, $|v| \geq 1$ such that $uv^iw \in L, \, \forall i \in \mathbb{N}$.
Let $n \in \mathbb{N}$ and $z = 0^p$ with $p \geq n \iff |z| \geq n$. Then, using the PL: $$ \exists u,v,w \quad \text{such that} \quad 0^p = uvw, \, |v| \geq 1 \rightarrow v = 0^m, \, 1 \leq m \leq n. $$
Since, $uv^iw \in L, \, \forall i \in \mathbb{N}$, this implies that each term of the sequence: $$ a_i = |uv^iw|=p + i \cdot m, \, i \in \mathbb{N} $$ is a prime. Is there a quick way to show that this is false (contradiction)?