Clearly $p=2$ works with solutions $(0,0)$ and $(1,1)$ and not $(0,1)$ or $(1,0)$. If $p \equiv 3(\bmod 4)$ then -1 is not a quadratic residue, so for $x^{3}+4x \neq 0$, exactly one of $x^{3}+4x$ and $-x^{3}-4x$ is a square and gives two solutions (for positive and negative $y$), so there's exactly two solutions for each such pair $\{x,-x\}$. If $x$ is such that $x^{3}+4x=0$, there's exactly one solution. If $p \equiv 1(\bmod 4)$, let $i$ be a square root of $-1(\bmod p)$. The right hand side factors as $x(x+2i)(x-2i)$. For $x=0,2i,-2i$ this is zero, there is one choice of $y$, namely zero. Otherwise, the right hand side is nonzero. For any fixed $x$, there are either 0 or 2 choices for $y$. Replacing $x$ by $-x$ negates the right hand side, again producing two choices for $y$ since -1 is a quadratic residue. So the total number of solutions $(x, y)$ is $3(\bmod 4)$, and thus there cannot be exactly $p$ solutions.