By induction on $n$ it is easy to show that $a_n \ge n$ for any $n$. Suppose that there exists a positive integer $k$ such that $a_k > k$. Let's choose such positive integer $m$ that $m : 2021!$ and $m > k$. Then for any $i = 2, 3, \dots, 2021$, $\text{GCD}(m, m + i) > 1$. It follows from the problem statement that $\text{GCD}(m, a_{m+1}) = 1$. Since $\{a_n\}$ is strictly increasing and $a_k > k$, then $a_{m+1} > m + 1$. Therefore, $m + 2 \le a_{m+1} \le m + 2021$, but then $\text{GCD}(m, a_{m+1}) > 1$ — a contradiction.

Let $b_n = a_n - n$ for each $n$. By induction on $n$ it is easy to show that $b_n \ge 0$ for any $n$. If $b_k > b_{k+1}$ for some $k$, then
$$a_k - k > a_{k+1} - k - 1 \implies a_k + 1 > a_{k+1} \implies a_k \ge a_{k+1}$$
— a contradiction. Thus, the sequence $\{b_n\}$ is non-decreasing. On the other hand, it has an upper bound: $b_n = a_n - n \le 2020$. Hence, there exists such non-negative integer $k$ and a positive integer $t$ that $b_n = k$ for each $n \ge t$. So, for any $n \ge t$
$$\begin{aligned} a_{n+1} \mid n^3 a_n - 1 &\implies n + k + 1 \mid n^3(n + k) - 1 \\ &\implies n + k + 1 \mid n^3(n + k) - 1 - (n + k + 1)(n^3 - n^2 + n(k + 1) - (k + 1)^2) = (k + 1)^3 - 1 \\ &\implies n + k + 1 \mid (k + 1)^3 - 1. \end{aligned}$$
But this is only possible when $k = 0$. Therefore, $b_n = 0$ for each sufficiently large $n$, and thus for all $n$, i.e. $a_n = n$ for all $n$.