Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is

Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.

I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y = 2b-a$, we see that $x+y = 4b$. Substituting,
$$xy+\dfrac {x+y}{2}+y=0$$
$$2xy+x+3y=0$$
From this we see that $y|x$, so we can substitute $x = ky$ for some integer $k$
$2ky^2+ky+3y = 0$
$$k = \dfrac {3}{2y+1}$$
From here we get that $y \in \{-2, -1 , 1 \}$, and each of the cases can be checked individually.

Left side looks almost like $(2b+1)^2$, indeed adding $1$ we can write $(2b+1)^2=a^2+a+1$. It is known that right side can be a perfect square only in few cases - see Integral value of $n$ that makes $n^2+n+1$ a perfect square. 
Specifically, when $a>0$, the right side lies between two consecutive squares: $a^2<a^2+a+1<(a+1)^2,$
so it cannot be itself a square. Similarly for $a<-1$, we have $(a+1)^2< a^2+a+1 < a^2$, same reasoning applies. So it only remains to check $a=-1$.
