Prime number CFG and Pumping Lemma
Your proof is incomplete, there are many cases you don't cover. You are not ruling out, for example, $u = 0^{m/2-3}$, $v=0$, $x=00$, $y=00$, $z=0^{m/2-2}$.

A good hint that your proof is wrong is that all you're using about prime numbers is that they're odd¹. Odd numbers do form a context-free language.

The pumping lemma gives you $uvxyz=0^p \in L$ and for all $n\in\mathbb{N}$, $uv^nxy^nz = 0^{p+(n-1)k} \in L$, with $k = |vy| \ge 1$. Now you need to prove that the numbers $\{p + (n-1)k \mid n\in\mathbb{N}\}$ cannot all be prime.


   Look at the values of $p + (n-1)k$ modulo $k+1$ when $n$ varies from $1$ to $k+1$. Since $k$ and $k+1$ are co-prime, one of them has to be $0$ modulo $k+1$, meaning that it is a multiple of $k+1$ (and not $k+1$ itself because it's larger), so it is not prime. Depending on your mathematics background, you may formulate this argument using the Chinese remainder theorem or any other equivalent theorem.


¹  And you didn't account for $2$ being prime, but that's ok, it's enough to consider sufficiently long words, because the status of a finite number of words doesn't affect the context-freedom of the language.