Can someone please explain the contradiction when we take $a^2 + 5b^2$ to be equal to $1$.
After multiplying both sides by $2 - \sqrt{-5}$ and factoring at a $3$ I get $$3\left[\left(2-\sqrt{-5} \right) \gamma + 3\delta \right] = 2-\sqrt{-5}$$ Why does this imply $2-\sqrt{-5}$ is a multiple of $3$ and why is that a contradiction?
It is a multiple of $3$ since you can write it as $3$ times another element of the ring, specifically you wrote it is $3 \times (\left(2-\sqrt{-5} \right) \gamma + 3\delta)$ and $\left(2-\sqrt{-5} \right) \gamma + 3\delta$ is an element of the ring.
This is a contradiction because for any inetegrs $c,d$ you have that $3(c+d \sqrt{-5})$ is $3c + 3d \sqrt{-5}$ and as the representation in the form $x + y \sqrt{-5}$ with $x,y$ integers is unique it would follow that $3$ divides $2$ and $-1$ (as integers), which it does not.