Switzerland Selection Solution

Solution:
Since $S$ is non-empty, it contains some integer $a$. Since $1 \mid 2^{a}+1$, it follows that $1 \in S$ and thus $2^{1}+1=3 \in S$. We call a pair $(m, n)$ of positive integers $k$-valid if we have $m, n \in S$, $n|m| 2^{n}+1$ and $m / n$ is divisible by $k$ distinct primes. We will show inductively that for every $k \geq 1$ there is a $k$-valid pair.

Lemma. If $(m, n)$ is $k$-valid and $p$ is prime divisor of $m / n$, then $\left(p^{\alpha} m, p^{\alpha} n\right)$ is $k$-valid for all integers $\alpha \geq 0$.

Proof. Using LTE, we have $v_{p}\left(2^{p m}+1\right)=v_{p}\left(2^{n}+1\right)+1$ and hence $p n|p m| 2^{p n}+1$. Also, since $p n \in S$, we have $p m \in S$. Because also $p m / p n=m / n$, it follows that $(p m, p n)$ is $k$-valid. Repeating $\alpha$ times gives the desired result.

For $k=1$, take $(m, n)=(3,1)$. Now assume $(m, n)$ is $k$-valid for some $k \geq 1$ and let $p$ be a prime dividing $m / n$. For any integer $t \geq 1$ we have
$$0<v_{p}\left(2^{t n}+1\right)=v_{p}\left(2^{n}+1\right)+v_{p}(t)<n+\log _{p}(t)$$
Thus, the $k$ primes we know to divide $m / n$ make up at most a fraction
$$\prod_{p} p^{n+\log _{p}(t)}<(m / n)^{n} t^{k}$$
of $2^{t n}+1$. Using the lemma, we can pick $t$ arbitrarily large such that $(t m, t n)$ is still $k$-valid. It follows that $2^{t n}+1$ is divisible by a prime $q$ that does not already divide $m / n$, hence $(t m q, t n)$ is $k+1$-valid. We conclude the solution by picking a $N$-valid pair $(m, n)$ where $N$ is large enough to ensure $m / n$ is divisible by at least 2023 primes greater than 2023. We then use the lemma to make sure the powers of all these primes are in the prime factorisation of $m$ are at least 2023 .

Solution:
We start by defining the following sequence: $a_{1}=1$ and $a_{n+1}=2^{a_{n}}+1$. For the same reason as in the first solution we have $1 \in S$ and therefore $a_{n} \in S$ for all $n$, so it's enough to show that an integer as in the problem statement will divide one of the $a_{n}$. Now we also define $r_{n}=a_{n+1} / a_{n}$. If $r_{n-1}$ is an integer it's odd (because $a_{n}$ is odd) and we can factorise:
$$a_{n+1}=2^{a_{n}}+1=\left(2^{a_{n-1}}\right)^{r_{n-1}}+1^{r_{n-1}}=\left(2^{a_{n-1}+1}\right) \sum_{i=0}^{r_{n-1}-1}(-1)^{n}\left(2^{a_{n-1}}\right)^{i}=a_{n} \sum_{i=0}^{r_{n-1}-1}\left(1-a_{n}\right)^{i}$$
and therefore:
$$r_{n}=\sum_{i=0}^{r_{n-1}-1}\left(1-a_{n}\right)^{i}$$
is an integer as well. Since $r_{1}=3$ is an integer we therefore get by induction that all $r_{n}$ are integers and the equation above holds for all $n$. Now assume we have a divisor $d$ of $n$ and look at the equation mod $d$, we get $r_{n} \equiv r_{n-1}$, therefore:
$$d\left|r_{n} \Longleftrightarrow d\right| r_{n-1}$$
If we apply this to some prime $p$ we realise that as soon as $p$ divides $a_{n}$ it divides not only $r_{n-1}$ but also $r_{m}$ for all $m \geq n$, so $v_{p}\left(a_{n}\right)$ is increasing and will eventually be bigger than 2023. Now let $p \mid a_{n}$ and $k=v_{p}\left(r_{n}\right)$; since $v_{p}\left(a_{n+1}\right)=v_{p}\left(a_{n}\right)+k>k$ and because $p^{k} \mid r_{n}$ but $p^{k+1} \nmid r_{n}$ we also get: $p^{k} \mid r_{n+1}$ but $p^{k+1} \nmid r_{n+1}$ so $v_{p}\left(r_{n+1}\right)=k=v_{p}\left(r_{n}\right)$. Apart of $n+1$ this is also true (again by induction) for all $m>n$. So if there was only a finite amount of primes among the divisors of $a_{n}$, we would get that $r_{n}$ is constant for all $n$ bigger than some $n_{0}$, which would imply exponential growth, but $a_{n}$ grows much faster than exponential, so we can conclude that at some point a new prime will appear, which completes the proof.