$S$ isn't the set of all primes: $a_{3i+1} \equiv 1 \pmod{3}$, $a_{3i+2} \equiv 2 \pmod{3}$ and $a_{3i+3} \equiv 2 \pmod{3}$. Therefore, $3$ isn't in $S$. $S$ is infinite because $a_{n+1} = a_n(a_n + n)$. Take $n = 3p$ with $p$ a prime. If $p | a_{3p} \hspace{0.25cm} \forall p$ prime, then the set is obviously infinite. If $p \nmid a_{3p}$ for a certain $p$, then there will be a prime $q$ with the property that $q|a_{3p+1}$ and $q\nmid a_{3p}$ making it an infinite set. Proof: $a_{3p+1} = a_{3p}(a_{3p} + 3p)$. We have $gcd(a_{3p},(a_{3p}+3p) = gcd(a_{3p},3p)$ and as $3p \nmid a_{3p}$, we have $gcd(a_{3p}, 3p) = 1$ meaning $a_{3p+1}$ has (atleast) one prime factor more than $a_{3p}$ had, making it an infinite set.

Assume that $S$ is finite. since if $p|a_i \Rightarrow p|a_j$ for all $j \ge i$ and $a_{n+1}=a_n(a_n+n)$, after some point $a_n+n$ stops making "new" primes, but since any $a_n$ has finite number of prime divisors, and $n$ varies by natural numbers, by Kobayashi's theorem $\{a_n+n \}$ have infinitely many prime divisors, hence contradiction, therefore $S$ is Infinite @Below I think it is fine to quote it without proof

Woah, Kobayashi's theorem is so powerful! Thanks for introducing me to something new xd. Is this allowed to quote in international contests (as I feel like the proof must be super complex!)?

When the first part is a cakewalk but you miscalculate $a_5$ so you don't get the second part Part 1 ($S$ is infinite): Assume otherwise and say $S = \{p_1, p_2, \cdots, p_k \}$. Then there exists $m$ with $p_i \mid a_m \forall i = 1, 2, \cdots, k$. Now choose $b = a_{hp_1p_2\cdots p_k + 1}$ with $hp_1p_2\cdots p_k + 1 > m$. then $a_{hp_1p_2\cdots p_k + 2} = b(b+hp_1p_2\cdots p_k + 1)$. But $b+hp_1p_2\cdots p_k + 1$ is coprime to each of the $p_i$ and isn't $1$, so there exists a new prime dividing $a_{hp_1p_2\cdots p_k + 2}$, contradiction. $\square$ Part 2 ($S$ isn't the set of all primes): We claim that $3 \notin S$. Indeed, if $a_{3i+1} \equiv 1 \pmod{3}, a_{3i+2} \equiv 2 \pmod{3}, a_{3i+3} \equiv 2 \pmod{3}, a_{3i+4} \equiv 1 \pmod{3}$. But $a_1 \equiv 1 \pmod{3}$, SO $3 \nmid a_i \forall i $ .$\square$