We claim $(x,y)=(\pm 1,\pm 1)$ is the only solution. Note that if $(x,y)$ is a solution, so is $(\pm x,\pm y)$. Hence, we assume $x,y\ge 0$ in the sequel. Summing up, we find $x^2+y^4=2$, and thus $y^4\le 2$: $0\le y<\pi/2$. Now, using $x=\sqrt{2-y^4}$, we arrive at \[ \sin^2\bigl(\sqrt{2-y^4}\bigr) -\sin^2 y = y^4-1. \]Now, check that $\max\{\sqrt{2-y^4},y\}<\pi/2$ in the range we consider. In particular, $\sin(\cdot)$ is increasing. Notice that if $y^4>1$ that is $y>1$, then $\sqrt{2-y^4}>y$, that is $y^4+y^2-2<0$. This implies $y<1$, a contradiction. Likewise, $y^4<1$ implies $y>1$. Hence $y=1$ is the only solution (for $y>0$). This completes the problem.