Solution:
Let
$$P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0.$$
If we plug in $x = 2020$, all terms except $a_0$ will be even, so since $P(2020)$ is odd, $a_0$ must be odd. But then if we plug in any other even number for $x$, $P(x)$ will still be odd since all the terms except $a_0$ will still be even, so $P(x)$ cannot be $0$. Thus, $P(x)$ can have no even roots.

However, if we plug in an odd number for $x$, the term $a_i x^i$ in $P(x)$ has the same parity as the term $a_i \cdot 2021^i$ in $P(2021)$ for each $i = 1, \ldots, n$. Thus, the sum $P(x)$ has the same parity as the sum $P(2021)$. But $P(2021)$ is odd, so $P(x)$ must be odd as well, and so not $0$. Thus, $P(x)$ cannot have odd roots either, so it has no integer roots.