Macedonian Mathematical Olympiad

Let $d=\text{gcd}(x,y)$. Then $d|\text{lcm}(x^2,y^2)$, $d|x^2$ and $d|y^2$, from where we get $d=1$. The equation is transformed into $1+x^2+y^2=x^2{y}^2$, or, equivalently, $(x^2-1)(y^2-1)=2$. We get $x^2-1=1$, $y^2-1=2$ or $x^2-1=2$, $y^2-1=1$, from where it follows that $x=2$, $y=3$ or $x=3$, $y=2$.