Find all solutions to the equation $x^2 + y^2 + xy = (xy)^2$ Can anyone help me find the number of solutions to the equation:

$$  x^2 + y^2 + xy = (xy)^2 $$

Let me give a brief account of what I've tried to proceed with:
Case 1: One of $x$ and $y$ is odd. This results into a contradiction where the parity of LHS and RHS differs.
Case 2: Both of $x$ and $y$ are even. I have shown that no solution apart from $x=y=0$ exists and I've proved that using Infinite Descent.
Now I'm stuck in case $3$ where both $x$ and $y$ are odd.

First, write it as $(x^2-1)(y^2-1)=xy+1$.
In the narrow ranges where $|x|\leq 1$ or $|y|\leq 1$, you can solve yourself.
Then, when $x>1$ and $x>1$ you have $(x^2-1)(y^2-1)>=2(x^2+y^2-2)=2(x-y)^2+4xy-4\geq 4xy-4$. 
But $xy\geq 4$, so $4xy -4> xy +3\cdot 4 -4 \geq xy-1$.
The case when $x<-1$ and $y<-1$ is the same.
There are no cases when $x<-1$ and $y>1$ because then $xy+1<0$.
Finally, if $x=0$, $y=0$. If $x=-1$ then $y=1$, and if $x=1,$ $y=-1$.
The key is that in most cases, it is "obvious" that $(x^2-1)(y^2-1)$ is a lot bigger than $xy$.

Another approach. Let $d=xy$. Then assume $d\neq 0$ and you have:
$$x^2+\frac{d^2}{x^2}=d^2-d$$
or
$$x^4-(d^2-d)x^2 + d^2=0$$
So by the quadratic formula:
$$x^2=\frac{d^2-d \pm \sqrt{d^2(d-3)(d+1)}}{2}$$
When is $(d-3)(d+1)=(d-1)^2-4$ a perfect square? The only case of perfect squares that differ by four is $0$ and $4$. That means $d=3$ or $d=-1$.
The only cases left to handle then are $d=0$ and $d=3$.
