Proof that $\{0^i1^{i^2}:i\in\mathbb{N}\}$ is not context free
The simplest way to show that your language is not context-free is by reduction to unary alphabet. Suppose that your language were context-free. Then the language obtained by applying a homomorphism which deletes all $0$'s would also be context-free. This language is the unary language $L' = \{ 1^{i^2} : i \notin \mathbb{N} \}$. It is known that a unary language is context-free iff it is regular. You can prove that $L'$ isn't regular in any number of ways (for example, it is finite but its limiting density is zero).

If you are hell-bent on using the pumping lemma, then the following argument works. As you mention, the only interesting case is Case 3, in which $x = 0^i$ and $z = 1^j$. According to the pumping lemma, for every $t$ the word $ux^{1+t}yz^{1+t}v = 0^{p+ti} 1^{p^2+tj}$ must belong to $L$, that is, $p^2+tj = (p+ti)^2$. But $(p+ti)^2 = \Theta(t^2)$ whereas $p^2+tj = \Theta(t)$, so for large $t$ the two cannot be equal.