Given natural $m$ and $n$, if $m^{2023}+2024m+n^2$ is divisible by $mn$, and $m>n$, then $m$ is a perfect square

Question

Given $m, n$ natural numbers, $m^{2023}+2024m+n^2$ is divisible by $mn$, and $m>n$, prove that $m$ is perfect square.

I tried considering $m$ being a multiple of $3$, $1+$ a multiple of $3$, and $2+$ a multiple of $3$, but I do not know how to continue.

Can someone give me an idea? Thank you in advance for your time.

Answer 1

There is a counterexample: when $m=3^2\cdot253$ and $n=3\cdot253$, it gives us $m^{2023}+2024m+n^2=9^{2023}\cdot253^{2023}+81\cdot253^2$ which is divisible by $27\cdot253^2=mn$. But, $m=3^2\cdot253$ is not a perfect square.