Determine all solutions $(x, y) \in \mathbf{N}^{2}$ of the equation:

$$
x(x+1)=4 y(y+1)
$$
$(x, y)=(0,0)$ is obviously a solution. Furthermore, the equation can be written as: $(2 y+1)^{2}=$ $4 y(y+1)+1=x^{2}+x+1$, for $x>0, x^{2}+x+1$, strictly between $x^{2}$ and $(x+1)^{2}$, cannot be a perfect square.