I am currently working on a special function defined by myself and found a special case that shows if $q \in \mathbb{N}$, then$$q(q(x-1)+1)^m = x^m$$ has no non-trivial natural integer solution $x$.
Although I have not extended my work over $\mathbb{Z}$, I believe it also has no solution in $\mathbb{Z}$. Can someone help verify this since the function I'm working on is a bit blurry for me right now.
P.S. Not sure if the equation is significant though.
It is easy to see what happens when $x=0$. Otherwise $q(x-1)+1 \neq 0$ and hence
$$\sqrt[m]{q}=\frac{x}{q(x-1)+1} \in \mathbb Q$$
This shows that $\sqrt[m]{q} \in \mathbb Z$, and hence $q=k^m$ for some integer.
Then, taking into account that $m$ could be even: $$k(k^m(x-1)+1) = \pm x \Leftrightarrow \\ k^{m+1}x-k^{m+1}+k = \pm x \Leftrightarrow\\ x(k^{m+1}\pm 1)=k^{m+1}-k $$
Now, this implies $k^{m+1}\pm 1|k^{m+1}-k$ we get that $$k^{m+1}\pm 1| (k^{m+1}\pm 1)-(k^{m+1}-k)=k\pm 1$$ and hence $$|k^{m+1}-1| \leq |k-1| \Rightarrow |k|^{m+1} \leq |k|+2$$
This leads to $|k|=0,1$ or $m=1$, from which the problem is easy to finish.