Since $LHS$ is a positive integer, $x-y$ is a perfect square, so let's set $x=y+t^2$ The equation becomes $2y^2+2t^2y+t^4+1089=2010|t|$ Since $t^2=|t|^2$ and $t^4=|t|^4$, we can assume that $t$ is a POSITIVE integer. $2y^2+2t^2y+t^4-2010t+1089=0$........ here the discriminant needs to be a perfect square, so $-4t^4+8\cdot 2010t-8 \cdot 1089= k^2$ so $4|k^2$.......let's write $k^2=4m^2$ and thus we have $-t^4+4020t-2178=m^2>0$ => $4020t>t^4+2178$ Notice that for $t>15$, this inequality cannot hold so $t \in [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]$ By checking, we notice that only $t=3$ makes the discriminant a perfect square and thus the answer is $(x,y)=(54,45)$