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

Multiply by $4$ and complete the square on both sides. This gives

$(4b+2)^2-4=(2a+1)^2-1$

$(4b+2)^2-(2a+1)^2=3$

What are the only two squares differing by exactly $3$?

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 $a=-1$.

Hmmm...

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

Where do $x$ and $y$ come from? Is the following what was meant?

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

I'll proceed along the lines of the latter. Clearly $a$ must be even, because if it's odd then $a^2 + a$ is singly even but $4b^2 + 4b$ is clearly doubly even. Therefore, with $a$ even, it follows that $$\frac{a^2 + a}{4}$$ is an integer, and so the equation can be rephrased as $$\frac{a^2 + a}{4} = b^2 + b.$$ Therefore $\sqrt{a^2 + a} = 2 \sqrt{b^2 + b}$. Trouble is, this requires both $\sqrt{a^2 + a}$ and $\sqrt{b^2 + b}$ to be integers.

Okay, I'm sorry, this is looking to be much more complicated than what you have...