Let $k=2021$. In fact, we will prove the statement for any integer $k > 0$.

Lemma 1 There exists $C > 0$ independent of $k$, such that $\tau(m) \le C m^{\frac{1}{k+1}}$ holds for all positive integer $m$, where $\tau(m)$ is the number of positive factors of $m$.

Proof of lemma 1 Let $m = p_1^{\alpha_1} p_2^{\alpha_2} \dots p_r^{\alpha_r}$ be the prime factorization of $m$. Evidently,
$$\tau(m) = (\alpha_1 + 1)(\alpha_2 + 1)\dots(\alpha_r + 1).$$
Equivalently, there exists $C > 0$ such that
$$\prod_{i=1}^{r} \frac{(\alpha_i + 1)^{k+1}}{p_i^{\alpha_i}} \le C^{k+1}$$
holds for all prime $p_i$ and positive integers $\alpha_i$. Note that polynomials grow slower than exponential functions, and hence for each prime $p \le 2^{k+1}$, the fraction $\frac{(\alpha + 1)^{k+1}}{p^{\alpha}}$ has an upper bound when $\alpha \ge 0$. Since the number of primes $p \le 2^{k+1}$ is finite, there exists $C'$ such that for any prime $p \le 2^{k+1}$,
$$\frac{(\alpha + 1)^{k+1}}{p^{\alpha}} < C'.$$
On the other hand, for $p > 2^{k+1}$,
$$\frac{(\alpha + 1)^{k+1}}{p^{\alpha}} < \left( \frac{\alpha + 1}{2^{\alpha}} \right)^{k+1} < 1.$$
It follows that $\prod_{i=1}^{r} \frac{(\alpha_i + 1)^{k+1}}{p_i^{\alpha_i}} \le C'^s$, where $s$ is the number of distinct prime factors of $m$ less than $2^{k+1}$. The existence of $C > 0$ is now verified.

Lemma 2 For any fixed integer $L > 0$, there exists $M > 0$ such that for any positive integer $m \ge M$, one can find a prime power $p^\alpha|m$, and $p^\alpha > L$.

Proof of lemma 2 Consider all prime powers less than or equal to $L$: there are only finitely many of them. Define $M$ equals the product of them plus $1$. If $m \ge M$ and $p|m$, then its highest power in $m$ satisfies $p^\alpha|m$, and $p^\alpha > L$.

Lemma 3 There exists a positive integer $D$ such that $a_n < 2n + D$ for every positive integer $n$.

Proof of Lemma 3 We take $D$ satisfying $D > \max_{1 \le s \le k}\{a_s\}$ and
$$D > 2^k C^{k+1} + C D^{\frac{k}{k+1}},$$
where $C > 0$ is defined in Lemma 1, and use induction to prove $a_n < 2n + D$ for every $n$. By the choice of $D$, $a_n < 2n + D$ is true for $n = 1, 2, \dots, k$. Assume $a_n < 2n + D$ holds for $1, 2, \dots, n + k - 1$, where $n$ is a positive integer. Now for $a_{n+k}$, it is known from the problem that $a_{n+k}$ is the smallest positive integer not dividing $a_n a_{n+1} \dots a_{n+k-1}$, and different from $a_1, \dots, a_{n+k-1}$. Hence,
$$a_{n+k} \le \tau(a_n a_{n+1} \dots a_{n+k-1}) + n + k.$$
From Lemma 1 and induction hypothesis, we have
$$\begin{align*} a_{n+k} &\le C(a_n a_{n+1} \dots a_{n+k-1})^{\frac{1}{k+1}} + n + k \\ &< C(2(n+k-1) + D)^{\frac{k}{k+1}} + n + k \\ &< C(2^{\frac{k}{k+1}}(n+k)^{\frac{k}{k+1}} + D^{\frac{k}{k+1}}) + n + k. \end{align*}$$
In the last step, we use the fact that $x^{\frac{k}{k+1}}$ is concave down for $x > 0$.

Now, if $n + k \le 2^k C^{k+1}$, then
$$\begin{align*} a_{n+k} &< C 2^{\frac{k}{k+1}} (n+k)^{\frac{k}{k+1}} + C D^{\frac{k}{k+1}} + n + k \\ &\le 2^k C^{k+1} + C D^{\frac{k}{k+1}} + n + k \\ &< D + 2(n+k); \end{align*}$$
if $n + k > 2^k C^{k+1}$, then
$$C 2^{\frac{k}{k+1}} (n + k)^{\frac{k}{k+1}} < n + k, \quad C D^{\frac{k}{k+1}} < D,$$
and
$$a_{n+k} \le n + k + D + n + k = 2(n + k) + D.$$
Therefore, $a_n < 2n + D$ is true for $n + k$.

Return to the original problem. For integer $m > 0$, if $m$ does not appear in the sequence, then, as the elements in $\{a_n\}$ are distinct, there must exist a positive integer $N_1$, such that $a_n > m$ for $n > N_1$. By definition of $\{a_n\}$, $m|a_n a_{n+1} \dots a_{n+k-1}$ for all $n > N_1$.

According to Lemma 2, there exists $M > 0$, such that for any integer $m \ge M$, we can find a prime power $p^\alpha|m$, and $p^\alpha > (3k)^k$. We claim that such $m$ must appear in $\{a_n\}$. Suppose to the contrary that for some $m > M$, $m$ does not appear in $\{a_n\}$. Fix a prime power $p^\alpha|m$, $p^\alpha > (3k)^k$. By the previous argument, there exists a positive integer $N_1$, such that for $n > N_1$,
$$m|a_n a_{n+1} \dots a_{n+k-1}.$$
For this $n$, there exists $i$, $n \le i \le n + k - 1$ and $p^{\lceil \alpha/k \rceil} | a_i$. Let
$$N_2 > 3N_1 + D + 3k,$$
and consider the set $A = \{n : 1 \le n \le N_2, p^{\lceil \alpha/k \rceil} |a_n\}$. On one hand, from Lemma 3, we have $a_n < 2N_2 + D$, and thus
$$|A| \le \frac{2N_2 + D}{p^{\lceil \frac{\alpha}{k} \rceil}} < \frac{2N_2 + D}{3k}.$$
On the other hand, since $m$ does not appear in $\{a_n\}$, $m|a_n a_{n+1}\dots a_{n+k-1}$ for $n > N_1$. Consequently, for any $k$ consecutive terms in $\{N_1 + 1, N_1 + 2, \dots, N_2\}$, at least one term belongs to $A$, implying that $|A| \ge \frac{N_2 - N_1}{k} - 1$. However, by the choice of $N_2$, we have
$$|A| \ge \frac{N_2 - N_1}{k} - 1 > \frac{2N_2 + D}{3k},$$
which contradicts the previous estimate. This means all integers $m \ge M$ must appear in $\{a_n\}$. $\Box$