$x(x+1) = y(y+4)$ <=>$4x^2+4x=4y^2+16y$ <=>$(2y+4)^2-(2x+1)^2=15$ <=>$(2y-2x+3)(2y+2x+5)=15=1*15=3*5$ <=>$2y-2x+3=1$ and $2y+2x+5=15$ or $2y-2x+3=3$ and $2y+2x+5=5$ <=>$x=3$ and $y=2$

$x^2+x=y^2+4y=(y+2)^2-4$. Thus, $x^2+x+4=(y+2)^2=(x+\frac{1}{2})^2+\frac{15}{4}$. So $(y+2)^2-(x+\frac{1}{2})^2=(y-x+\frac{3}{2})(y+x+\frac{5}{2})=\frac{15}{4}$. Since $y-x$ and $y+x$ are integers, when expressed in simplest form, both $y-x+\frac{3}{2}$ and $y+x+\frac{5}{2}$ will have denominator $2$. So their numerators must multiply up to $15$. The numerators are both positive because $y+x+\frac{5}{2}$ is positive. Since $y-x+\frac{3}{2}<y+x+\frac{5}{2}$, we have two cases. Case 1: $y-x+\frac{3}{2}=\frac{1}{2}$ and $y+x+\frac{5}{2}=\frac{15}{2}$. Then $y-x=-1$ and $y+x=5$. This implies $\boxed{x=3,y=2}$, which works. Case 2: $y-x+\frac{3}{2}=\frac{3}{2}$ and $y+x+\frac{5}{2}=\frac{5}{2}$. Then $y-x=0$ and $y+x=0$, a contradiction.

Obviously, $x-3<y<x$. If $y=x-2$, $x(x+1)=(x-2)(x+2)\iff x=-4<0$. If $y=x-1$, $x(x+1)=(x-1)(x+3)\iff x=3,y=2$.