Well

The real answer is in fact 1. I will prove this through infinite descent. Clearly, $(0, 0)$ is a solution. Now, assume there exists a solution with $|x|+|y|>0$ which is minimal. Consider $x^2 = 5y^2$ Implying $5|x$ . Let's replace $x$ with $5x_1$ . Substituting, we get $25x_1^2 = 5y^2$ $\Rightarrow 5x_1^2=y^2$ Implying $5|y$ . Substituting $y=5y_1$ , we get $5x_1^2=25y_1^2$ $x_1^2=5y_1^2$ Notice that this implies $(x_1, y_1)$ is also a solution, with $|x_1|+|y_1|>\dfrac{|x|+|y|}{5}$ . But we already assumed $x, y$ was the minimal solution! Thus, by infinite descent, $(0, 0)$ is the only solution.