Very nice. Set $b_n\triangleq a_n-n$, and note that $b_n\ge 0$ is a sequence of integers. Now the conditions yield a) $b_n$ is bounded from above by $2020$, and b) $a_{n+1}>a_n\Leftrightarrow b_{n+1}>b_n-1\Rightarrow b_{n+1}\ge b_n$. Namely, $\{b_n\}_{n\ge 1}$ is a non-decreasing sequence of non-negative integers, bounded from above. Consequently, $b_n$ is eventually constant: there exists an $N\in\mathbb{Z}^+$ and $c\ge 0$ such that $b_n = c$ for every $n\ge N$. Now, we return the divisibility relation. In particular, we investigate this for $n\ge N$: \[ a_{n+1}\mid n^3 a_n -1 \Rightarrow n+1+b_{n+1}\mid n^4 + n^3b_n -1 \Rightarrow n+1+c \mid n^4+n^3c-1. \]Now $n+1+c\mid n^4+n^3c+n^3$ as well. Hence, $n+1+c\mid n^3+1=(n+1)(n^2-n+1)$. Next, note that $n+1\equiv -c\pmod{n+1+c}$, and $n^2-n+1 \equiv (n+1)^2-3(n+1)+3 \equiv c^2+3c+3\pmod{n+1+c}$. Thus, \[ n^3+1\equiv 0\pmod{n+1+c}\Leftrightarrow c(c^2+3c+3)\equiv 0\pmod{n+1+c}. \]We are almost done: we have $n+1+c\le c(c^2+3c+3)$ unless $c\ne 0$. Sending $n\to\infty$ yields a contradiction unless $c=0$. Thus we obtain $c=0$ eventually. But since $\{b_n\}_{n\ge 1}$ is a non-decreasing sequence of non-negative integers, it therefore follows that $b_n=0$ for all $n$, namely $a_n=n$ for all $n$, as desired.

It's easy to get $a_{n}\ge{\dots}\ge{a_{1}+(n-1)}\ge{n}$ We're given $a_{n+1}\mid{n^3.a_{n}-1}$ , for all $n\in{\mathbb{N}}$ (1) Now , for the sake of contradiction , assume $\exists{i\in{\mathbb{N}}}$ for which $a_{i}>i$. (1)$\implies{\gcd(a_{n+1},n)=1}$ , for all $n\in{\mathbb{N}}$ (2). And if we take $j$ such that $2021!\mid{j}$ and $j>i$ , then $\gcd(j,j+t)>1$ for all $2\le{t}\le{2021}$ . (3) Since the sequence is increasing and $a_{i}>i$ and $j>i$ , we have $a_{j+1}>j+1$ Then , $j+2\le{a_{j+1}}\le{j+2021}$ and so $1\overset{(2)}{=}\gcd(a_{j+1},j)\overset{(3)}{>}1$ , contradiction . $\blacksquare$

Clear $a_1=1$, then $a_n\geq n$ due to $\{a_n\}$ is strictly increasing . Only to consider $a_n=b_n+n$ due to $\{b_n\}$ is bounded and $b_n\geq 0$. Now star with $n+1+b_{n+1} \mid n^4+n^3b_n-1$, $\text{mod} (n+1+b_{n+1})$ tell us that $$n+1+b_{n+1} \mid (b_{n+1}+1-b_n)(b_{n+1}+1)^3-1.$$Let $n \to\infty$ and notice $\{b_n\}$ is bounded, we get $(b_{n+1}+1-b_n)(b_{n+1}+1)^3-1=0$, then $b_{n+1}=0$ for large enough $n$. Clear $a_1=1$, and $\{b_n\}$ is strictly increasing we know $b_n=0$ for all $n$.

Define $Q_i=p_1p_2 \dots p_i$, where $p_i$ is the $i$-esime prime. Note that $a_{n+1}$ and $n$ are coprimes, so $a_{Q_i+1}$ and $Q_i$ are coprimes. Suppose that $a_{Q_i+1}>Q_i+1$ (obviously $a_n \geq n$) and note that $Q_i+2$, $Q_i+3$, $\dots Q_i+p_{i+1}-1$ have at least one divisor prime $p_1$, $p_2$, $\dots p_i$. Conclude that $a_{Q_i+1}\geq Q_i+p_{i+1}$. Then, it's sufficient to consider $i$ such that $p_i $ is greater than $2020$ to get a contradiction. Finally $a_n=n$

$\mathbf{Statement:}$ If there is some $a_i$ for which $a_i\geq i+j$ then there is some $a_s$ that $a_s\geq s+j+1$. $\mathbf{Proof:}$ Let's consider any $w\geq i$, then for it it will be that $a_w\geq w+j$, now let's pick up $w$ such that $(j+1)|w$ and consider the fact that $a_{w+1}|w^3a_w-1$, and we also know that $a_{w+1}\geq w+j+1$, consider two cases... The first, $a_{w+1}=w+j+1$, and it is divisible by $j+1$, so $w^3a_w-1$ is divisible by $j+1$, which is impossible because $w$ is divisible by $j+1$, respectively, this case is impossible. The second, $a_{w+1}\geq w+j+2$, means we have proved our statement. $\mathbf{Conclusion:}$ It turns out that each time we will find some $a_n\geq n+s$ where each time $s$ will increase for some large $n$, which will give a contradiction on the point that $a_n\leq n+2020$.

wer wrote: Nice one .. are you about the problem or my solution?

By strictly increasing $a_i$ we get $0 \le a_n-n \le 2020$. Also notice that $a_n-n$ is non-decreasing. So, $a_n-n$ is eventually constant for all $n \ge c$. Lets say that $a_n-n = M$ for $n \ge c$. So, for any $n \ge c$ we have $a_{n+1}=a_n+1 | n^3a_n-1 \implies a_n+1|n^3 +1 \implies n+M+1|n^3+1$. If $M+1 > 1$ then we set a $n \ge c$ such that $M+1|n$. So, $M+1|n+M+1|n^3+1$ which yields a contradiction so $M=0$. Hence, $a_n=n$ as desired.