Problem: Determine whether there exists a function $f \colon \ZZ_{> 0} \to \ZZ_{>0}$
such that for all positive integers $m$ and $n$,
\[ f(m+nf(m))=f(n)^m+2024! \cdot m. \]

Solution: The answer is no. Let $P(m,n)$ denote the given FE.

\paragraph{Solution 1 (Gopal Goel).}
Suppose there was a function $f$, and let $r=f(1)$. Note that $P(1,n)$ gives
\[ f(1+rn) = f(n)+2024!. \]
Iterating this result gives
\[ f(1+r+\dotsb+r^k) = r+k\cdot 2024! \]
for all $k \in \ZZ_{\ge 0}$.
If $r=1$, this implies $f(k+1) = 1 + k\cdot 2024!$, which isn't a valid solution, so $r\ge 2$.

Let $m = 1 + r + \dotsb + r^{r^2(r-1)}$. Note that
\[ 1+r+\dotsb+r^k = m + nf(m) \iff \frac{r^{k+1}-r^{r^2(r-1)+1}}{r-1} = n[r+r^2(r-1)2024!]. \]
This has a positive integer solution for $n$ as long as $k$ is sufficiently large in terms of $r$ and
\[ k\equiv r^2(r-1) \pmod{\phi(1+r(r-1) \cdot 2024!)}. \]
For such $k$, $P(m,n)$ implies
\[ r + k\cdot 2024! = f(n)^m + 2024!\cdot m. \]
In particular, $r + (k-m)\cdot 2024!$ is a perfect $m$th power.
This has to be true for all $k$ sufficiently large in an arithmetic progression,
which is clearly impossible, as desired.

\paragraph{Solution 2 (Carl Schildkraut).}
As in the previous solution, we have
\[ f(1+r+\dotsb+r^k) = r+k\cdot 2024! \]
for all $k \in \ZZ_{\ge 0}$, where $r=f(1)\ge 2$.

\begin{claim*}
  We have $f(a)\ge\frac{2^a-1}{2024!\cdot a}$ for $a\in\mathbb{Z}_{>0}$.
\end{claim*}
\begin{proof}
  Let $c = a+af(a)$, $b = f(c)+a$, and $d=f(a)$, so that
  \[ a + bf(a) = a + af(a) + f(a)f(c) = c + df(c). \]
  Comparing $P(a,b)$ and $P(c,d)$ tells us
  \[ f(b)^a + 2024!\cdot a = f(d)^c + 2024!\cdot c, \]
  so
  \[ f(b)^a - \left(f(d)^{1+f(a)}\right)^a = 2024!\cdot af(a). \]
  Since the left side of the above equation is positive,
  it must be at least $2^a-1$, which implies the claim.
\end{proof}
Plugging in $a=1+r+\dotsb+r^k$ into the above claim immediately
gives the desired contradiction, for sufficiently large $k$.

\paragraph{Solution 3 (students).}
If $f$ existed, then for any $k \in \ZZ_{>0}$ we would have
\begin{align*}
  f(f(3))^{3+kf(3)} + 2024!(3+kf(3)) &= f(3+kf(3)+f(3)f(3+kf(3))) \\
  &= f(3+(k+f(3+kf(3))f(3))) \\
  &= f(k+f(3+f(3)k))^3 + 2024!\cdot 3.
\end{align*}
Choosing $k=27(2024!)^2f(3)^2$ gives
\[ f\left(k+f(3+f(3)k)\right)^3 = \left(f(f(3))^{1+9(2024!)^2f(3)^3}\right)^3+ (3\cdot 2024!f(3))^{3} \]
which contradicts Fermat's last theorem (no two perfect cubes may sum to a perfect cube).
Hence $f$ cannot exist.