The emperor's new integer solutions

Find all pairs of integers $a$ and $b$ that are solutions to the following equation: $$a\cdot(a+1)\cdot(a+2) = b^2+4.$$

That’s it. That’s the riddle.

First, since $a$, $a+1$ and $a+2$ are three consecutive integers, one of them must be divisible by $3,$ so therefore their product $a \cdot (a+1) \cdot (a+2)$ is divisible by three. So regardless of the value of $a \in \mathbb{Z}$, the lefthand side of the equation is divisible by $3$.

Secondly, we can show that the righthand side of the equation cannot be divisible by three. If $3 \mid b,$ say with $b = 3\ell$, then clearly $$b^2+4 = 9 \ell^2 + 4 = 3(3 \ell^2 + 1) + 1 \equiv 1 \!\!\!\!\mod 3,$$ so if $3 \mid b$ then $3\not\mid b^2 + 4.$ On the other hand $3 \not\mid b,$ then Fermat's little theorem states that $$b^2 \equiv 1 \!\!\!\!\mod 3,$$ so for instance, if $3 \not\mid b$ then there is some $m \in \mathbb{Z}$ with $$b^2 = 3m + 1, so b^2 + 4 = 3m + 5 = 3(m+1) + 2 \equiv 2 \!\!\!\!\mod 3.$$ Thus, regardless of the value of $b \in \mathbb{Z}$, the righthand side of the equation is not divisble by $3.$

Therefore, since for any $a \in \mathbb{Z}$ the lefthand side is divisible by $3$, but for any $b \in \mathbb{Z}$ the righthand side cannot be divisible by $3,$ then there are no integer solutions to $$a\cdot(a+1)\cdot(a+2) = b^2 + 4.$$

While Fermat shuts the party down for $\mathbb{Z}^2,$ if we invite Gauss in, there are solutions over the Gaussian integers, e.g., $(-2, \pm 2i),$ $(-1, \pm 2i),$ and $(0, \pm 2i).$ But this might be a bridge too far for this particular riddle.