, 2021

We claim that $g(x) = x^3 - 2019x - 2021$ is irreducible over $\mathbb{Q}$. Since $\deg g = 3$, if $g$ were reducible, then it must have a linear factor. In other words, $g$ would have a rational root $c$. Since $g$ is monic, $c$ must be an integer by the rational root theorem. However, $c^3 - 2019c$ must be even, so it is impossible that $c^3 - 2019c - 2021 = 0$. This proves our claim.

Note that $\alpha$ is a root of $g(x)$. It follows from the above claim that $g$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ (see remark). From the given condition $g(f(\alpha)) = 0$, we know that $g(x) \mid g(f(x))$. Replacing $x$ by $f(x)$, we obtain $g(f(x)) \mid g(f^2(x))$. Since $g(x)$ divides $g(f(x))$, and $g(f(x))$ divides $g(f^2(x))$, we have $g(x) \mid g(f^2(x))$. Similarly, for any positive integer $n$, we have $g(f^{n-1}(x)) \mid g(f^n(x))$. By induction, this yields $g(x) \mid g(f^n(x))$. Therefore, $g(f^n(\alpha)) = 0$. This is exactly the same as $(f^n(\alpha))^3 - 2019f^n(\alpha) = 2021$ as desired.

Remarks. We say that $g$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$ if $g$ is the monic polynomial with rational coefficients of smallest degree such that $g(\alpha) = 0$. An important property of the minimal polynomial is that for any polynomial $h$, $h(\alpha) = 0$ if and only if $g$ divides $h$. Note that this also implies the minimal polynomial is the only irreducible monic polynomial having $\alpha$ as a root.
Here is a proof. For any polynomial $h$ such that $h(\alpha) = 0$, let
$$h(x) = g(x)q(x) + r(x)$$
where $\deg r < \deg g$. Then $h(\alpha) = g(\alpha)q(\alpha) + r(\alpha)$, which implies $r(\alpha) = 0$. Due to the minimal choice of the degree of $g$, $r$ must be the zero polynomial. This means $g(x) | h(x)$. Conversely, if $g(x) | h(x)$, then it is obvious that $h(\alpha) = 0$.