For $b=0$, $a=0$ works. Let $b>0$. Note that \[ a^2 = b^{2022}-b\implies \left(b^{1001}-1\right)^2 < b^{2002}-b < b^{2002} \]unless $b=1$. Hence, $b=1$ forcing $a=0$. The case, $b<0$, is handled analogously.

The equation is equivalent to $$a^2=b^{2022}-b$$So $b^{2022}-b$ is a perfect square $\Rightarrow b\geq 0$ For $b>1$, by bounding between squares we have $$(b-1)^{2022}<b^{2022}-b<b^{2022}$$Hence the only solution are $(0,0)$ and $(0,1)$

wehaventawake wrote: The equation is equivalent to $$a^2=b^{2022}-b$$So $b^{2022}-b$ is a perfect square $\Rightarrow b\geq 0$ For $b>1$, by bounding between squares we have $$(b-1)^{2022}<b^{2022}-b<b^{2022}$$Hence the only solution are $(0,0)$ and $(0,1)$ But you are not bounding between consecutive squares...

wehaventawake wrote: The equation is equivalent to $$a^2=b^{2022}-b$$So $b^{2022}-b$ is a perfect square $\Rightarrow b\geq 0$ For $b>1$, by bounding between squares we have $$(b-1)^{2022}<b^{2022}-b<b^{2022}$$Hence the only solution are $(0,0)$ and $(0,1)$ it should be $$(b^{1011}-1)^2<b^{2022}-b<b^{2022}$$