, 2023

Solution:
Assume for the sake of contradiction that there are infinitely many $f$-cool integers.
If $f(x)$ has a negative leading coefficient, then a sufficiently large $f$-cool integer $m$ will have $f(m)<0$. But this implies $m$ is not $f$-cool, contradiction.
Thus $f(x)$ has a positive leading coefficient, so we can pick an $N$ such that for all $m>N$,
$$f(m)>\max (f(1), f(2), \ldots, f(m-1))$$
This means $f(m)$ is the largest value in $f[\operatorname{divs}(m)]$, so if $m$ is $f$-cool with $f[\operatorname{divs}(m)]=\operatorname{divs}(n)$, then we must have $n=f(m)$, since $n$ is the largest value in $\operatorname{divs}(n)$. In other words,
$$f[\operatorname{divs}(m)]=\operatorname{divs}(f(m))$$
for all $f$-cool $m>N$.
For each of those $m$'s, $1 \in \operatorname{divs}(f(m))$, so there must be a $k \in \operatorname{divs}(m)$ such that $f(k)=1$. Let $k_{1}, k_{2}, \ldots, k_{n}$ be the solutions to $f(x)=1$. Thus every $f$-cool $m>N$ is divisible by some $k \in\left\{k_{1}, k_{2}, \ldots, k_{n}\right\}$. Since there are infinitely many such $m$'s and finitely many $k$'s, by the Pigeonhole Principle there is some $k$ which divides infinitely many $f$-cool integers $m$. (Note that $k \neq 1$ since $f(1) \neq 1$.)
For all $f$-cool $m>N$ divisible by $k$, we have
$$\left.f\left(\frac{m}{k}\right) \in f[\operatorname{divs}(m)]=\operatorname{divs}(f(m)) \Longrightarrow f\left(\frac{m}{k}\right) \right\rvert\, f(m) .$$
Thus, $f(x) \mid f(k x)$ has infinitely many positive integer solutions. Let $d=\operatorname{deg}(f)$, and write
$$\frac{f(k x)}{f(x)}=k^{d}+\frac{g(x)}{f(x)}$$
for some $g(x) \in \mathbb{Z}[x]$ with $\operatorname{deg}(g)<d$. If $g(x) \neq 0$, then for sufficiently large $x$ we have $0<|g(x)|<f(x)$, since $\operatorname{deg}(f)>\operatorname{deg}(g)$. But then $\frac{f(k x)}{f(x)}-k^{d}=\frac{g(x)}{f(x)}$ cannot be an integer, which gives us the desired contradiction.
Therefore $g(x)=0$, so $f(k x)=k^{d} f(x)$, i.e. $f(x)=a x^{d}$ for some positive integer $a$. If $a=1$ then $f(1)=1$, a contradiction. But if $a>1$, then $f(x)=1$ has no integer solutions, another contradiction.