Answer: D

Solution:

Move all terms to one side and complete the square: $$\begin{aligned} &x^{2020}+y^2=2y \\ &\quad \Longrightarrow x^{2020}+(y-1)^2=1. \end{aligned}$$ Because $(y-1)^2\ge 0,$ we must have $x^{2020}\le 1.$ Since $x$ is an integer, $x=-1,0,1.$

If $x=\pm1,$ then $(y-1)^2=0,$ so $y=1.$ If $x=0,$ then $(y-1)^2=1,$ so $y=0$ or $2.$ This gives $4$ ordered pairs.

Thus, D is the correct answer.