Perfect squares

Assume without loss of generality that $x \ge y$ .

Thus, we have that $x^2<x^2+3y \le x^2+3x <(x+2)^2$

So, we have that $x^2+3y$ is a square, squeezed between $x^2$ and $(x+2)^2$ .

So, it is in fact $(x+1)^2$ , and so $3y=2x+1$ . We may thus substitute $x$ to $3u-2$ and $y$ to $2u-1$ , where $u$ is a natural. Therefore we have that $y^2+3x=(2u-1)^2+3 \times (3u-2)=4u^2+5u-5$ And so $4u^2 \le y^2+3x <(2u+2)^2$

So $u$ is either $6$ or $1$ . So we have that $(x,y)=(11,16),(16,11),(1,1)$ The answer is $3$ .