SAMC

From $3\mid x^2+y^2$ it follows $3\mid x$ and $3\mid y$, hence $x=3u$ and $y=3v$, for some positive integers $u$ and $v$. Replace in the equation and get $3\left(u^2+v^2+11^2\right)=670\sqrt{3(u-v)}$, hence $u-v=3k^2$, for some positive integer $k$, i.e. $u^2+v^2+11^2=670k$. We have $u^2+v^2>(u-v{)}^2=9k^4$, and obtain $9k^4+121<670k$, hence $k<5$.

Case 1. If $k=2$ or $k=4$, then $u^2+v^2+11^2\equiv 0(\mathrm{mod}4)$, not possible since we have $11^2\equiv 1(\mathrm{mod}4)$ and $u^2,v^2\equiv 0$ or $1(\mathrm{mod}4)$.

Case 2. If $k=1$, then we get the system $$\{\begin{array}{l}{u}^2+v^2=549\\ u-v=3\end{array}$$ having integer solutions $u=18$ and $v=15$. In this case it follows $x=54$ and $y=45$.

Case 3. If $k=3$, then we get the system $$\{\begin{array}{l}{u}^2+v^2=1882\\ u-v=27\end{array}$$ having no solutions in integers.