Since x and y are positive, we may write -

$x^2+3y=(x+a{)}^2$, and
$y^2+3x=(y+b{)}^2$

where a, b are positive integers.

Expanding, we find that the squared terms cancel, leaving the linear simultaneous equations -
$3y=2ax+a^2$
$3x=2by+b^2$
Solving, we obtain -
$x=\frac{2a^{2}b+3b^2}{9-4ab}$
$y=\frac{2b^{2}a+3a^2}{9-4ab}$

Since a and b are positive, the numerator in each fraction will be positive. For the denominator to be positive, we must have ab = 1 or 2.

If (a,b) = (1,1), (1,2), (2,1), then, respectively, (x,y) = (1,1), (16,11), (11,16). Hence, these are the only solutions.