Trying to be pretty

Simplifying, we have $(x+y)^2=xy(1+xy), \hspace {5pt} x\neq 0, y\neq 0$ This implies a perfect square is equal to the product of two consective integers, which is impossible unless both sides equal to zero. Thus it follows that $x+y=0, xy+1=0$ This yields two solutions $x=1,y=-1$ and $x=-1,y=1$ .

Without loss of generality, assume $x^{2}\geqslant y^{2}$ .

Then $\frac{3}{y^{2}}\geqslant \frac{1}{x^{2}}+\frac{1}{xy}+\frac{1}{y^{2}}=1.$

Therefore $y^{2}\leqslant 3\: \Rightarrow y=\pm 1$ .

When $y=1$ , $x=-1$ ;

When $y=-1$ , $x=1$ .

$\therefore$ There are $\boxed{2}$ solutions.