Can we find pairs $(x,y)$ of positive integers such that $x^2+3y$ and $y^2+3x$ are simultaneously perfect squares? Thanks a lot in advance. My progress is minimal.
Not a complete solution but an approach that seems like it will work.
Assume $y \gt x$
Then we have that
$(y+2)^2 \gt y^2 + 3y \gt y^2+3x \gt y^2$
If $y^2 + 3x$ was a perfect square, then we have that $y^2 + 3x = (y+1)^2$.
This gives us $3x = 2y+1$.
Substitute in the other expression, and form similar inequalities. This will narrow down to few small cases to consider.
The other case $y=x$ can be treated similarly.