Finding the Integers!

Of course, $(1,1)$ is a solution. Suppose that $(x,y)$ is a solution with $x > 1$ .

If $p > 2$ is an odd prime factor of $x$ , let $a,b \in \mathbb{N}$ be the indices of $p$ in $x$ and $y$ respectively, so that $x = p^aq$ and $y = p^br$ for some integers $q,r$ which are coprime to $p$ . Comparing the indices of $p$ in $y^{x^2}$ and $x^{y+2}$ , we deduce that $bx^2 = a(y+2)$ . SInce $p$ is an odd prime factor of $y$ , $p$ and $y+2$ are coprime. Since $p^{2a}$ divides $x^2$ , it follows that $p^{2a}$ divides $a$ . This is impossible (since $p^{2a} > a$ ), so we deduce that $x$ has no odd prime factor, and hence $x$ is a power of $2$ . If $y$ had any odd prime factor, so would $x$ , so it also follows that $y$ is a power of $2$ .

Suppose then that $x = 2^u$ and $y = 2^v$ for positive integers $u,v$ . Then $2^{vx^2} \,=\, 2^{u(y+2)}$ , so that $vx^2 \,=\, u(y+2)$ , and so $v2^{2u} \,=\, u(2^v + 2)$ .

If $v \ge 2$ then $2^v + 2$ is even, but not divisible by $4$ , and hence $2^{2u-1}$ must divide $u$ . This is also impossible (since $2^{2u} > 2u$ ), so we deduce that $v = 1$ and $2^{2u} = 4u$ , and hence $u=1$ . Hence it follows that $(x,y) = (2,2)$ .

The $\boxed{2}$ solutions of this equation are $(1,1)$ and $(2,2)$ .