2020 AMC 12B Problem 8

Completing the square gives $x^{2020} + (y - 1)^2 = 1.$ Both terms are nonnegative, so $x^{2020} \le 1,$ forcing $x \in \{-1, 0, 1\}.$ If $x = 0,$ then $(y - 1)^2 = 1,$ giving $y = 0$ or $y = 2.$ If $x = \pm 1,$ then $x^{2020} = 1,$ so $(y - 1)^2 = 0$ and $y = 1.$ The solutions are $(0, 0), (0, 2), (1, 1),$ and $(-1, 1)$ — four in all. Thus, the correct answer is D.