Here is my construction, which doesn't involve any consideration of cases. @below, Citing Mihailescu wasn't worth any points as it trivializes the problem.

By Catalan/Mihailescu, we can take $M_n=\mathbb{N}$ for all $n \ge 4$ (and still essentially the same for $n=2$ and $n=3$). Even if you don't want to cite this, it helps to know that the answer definitely must be YES...

LET, quite easy.

The answer is $\boxed{\text{yes}}$. This problem is trivialized by Catalan's Theorem which states that the only consecutive perfect powers are $2^3$ and $3^2$. Because of this we can just take $m\geq 4$.

Here is my solution only using Lte We can clearly see that if $n\geq 4\;$ $n^2-1$ will have atleast 2 distinct prime factor let it be $p,q$ such that $v_{p}(n^{^{2}}-1)=x$ and $v_{q}(n^{^{2}}-1)=y$ where $x,y \in \mathbb{N}$ let $r$ be a prime which $gcd(r,y)=1,~r>x$ and $M_{n}= \{2(p^{r-x}+p^r(q)i)|i\in \mathbb{N}\}$ from Lte $ \rightarrow v_{p} (n^m-1)$ $=$ $v_{p} (n^2-1)+v_{p} (p^{r-x}+p^r(q)i)=x+(r-x)=r ~~ \forall m\in M_{n}$ $and ~ v_{q} (n^m-1)=v_{q} (n^2-1)+v_{q} (p^{r-x}+p^r(q)i)=y+0=y ~~ \forall m\in M_{n}$ $there~for~ gcd(v_{p}(n^m-1), v_{q}(n^m-1))=gcd(r,y)=1~~ \forall m\in M_{n} $ $\therefore ~$ $n^m-1$ isn't a perfect power $\forall m\in M_{n}$ $for~n=2$ we can choose $M_{2}=\{2+6i| i\in \mathbb{N}\}$ thus $v_{3}(n^m-1)=1 ~~\forall m\in M_{2}$ $for~n=3$ we can choose $M_{3}=\{1+2i| i\in \mathbb{N}\}$ thus $v_{2}(n^m-1)=1 ~~\forall m\in M_{3}$ $\therefore ~$ for every $n\in \mathbb{N}-\{1\}$ there exist $M_{n}$ which $n^m-1$ isn't a perfect power $\forall m\in M_{n}$

VicKmath7 wrote: Here is my construction, which doesn't involve any consideration of cases. @below, Citing Mihailescu wasn't worth any points as it trivializes the problem.

In the lte part is there a reason the font got smaller it also hapens with me do you know how to fix it im new to latex