Solution:

Move all terms to one side and complete the square: $$x^{2020}+y^2=2y\quad\Longrightarrow\quad x^{2020}+(y-1)^2=1.$$ 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.