How to prove $(2^m-1)\nmid(2^n+1)$ if $m>2$ Let $m,n$ be positive integers, and $m>2$, show that
$$(2^m-1)\nmid(2^n+1)$$
if $m=n$ it is clear hold, because $\dfrac{2^n+1}{2^n-1}=1+\dfrac{2}{2^n-1}\notin N$ 
and if $m>n$, then we have 
$$2^m-1\ge 2^{n+1}-1>2^n+1$$
then this case also hold. But $m<n$ it seems hard to prove it
and maybe this problem have other methods,Thanks
 A: We'll use the following result a few times:
${\;\;\;\;\;\;}$ Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ 
Suppose that $2^m-1\mid2^n+1$.
Then $$\begin{align*}2^m-1\mid2^{2n}-1&\implies\gcd(2^m-1,2^{2n}-1)=2^m-1\\
&\implies2^{\gcd(m,2n)}-1=2^m-1\\
&\implies \gcd(m,2n)=m\\
&\implies m\mid2n.\qquad(1)\end{align*}$$
On the other hand
$$\begin{align*}&\gcd(2^m-1,2^n-1)\mid\gcd(2^n+1,2^n-1)=\gcd(2^n+1,2)=1\\
&\implies2^{\gcd(m,n)}-1=1\\
&\implies\gcd(m,n)=1.\qquad(2)\end{align*}$$
Combining $(1)$ and $(2)$ gives $m\mid2$.
Note: In exactly the same way we can prove that $3^m-1\mid3^n+1\implies m\mid2$. For $a>3$, $a^m-1\mid a^n+1$ is impossible because the RHS is not divisible by $a-1$.
Side remark: You may note that the deduction of $(1)$ generalizes to:
${\;\;\;\;\;\;}$ Showing that $a^n - 1 \mid a^m - 1 \iff n \mid m$
A: The quesion is whether it is possible that $$ 2^m-1 | 2^n +1 $$
Of course, in general this requires that $m \lt n$ (except possibly $m,n$ small numbers, see below for this).       
So let $n=am+r$  where  $0 \le  r \lt m$ .     
Then we can derive     
$\qquad  2^{am+r} +1 = 2^{am}2^r +1 \\
\qquad \qquad \qquad = (2^{am}-1)2^r +2^r+1 \\
\qquad \qquad \qquad = (2^{am}-1)2^r +(2^r+1) $        
It is well known, that $2^{am}-1$ is divisible by $2^m-1$ , so if indeed $(2^m-1) | (2^n-1)$ we have        
at one hand that $(2^m-1) | (2^{am}-1)$        
and on the other hand it must thus also be true, that $2^m-1 | 2^r+1$      
But this can only be true if $m=2,r=1$ and $2^2-1 = 3 |2^1+1 = 3 $ and never else because $r<m$ by definition.
A: Deny. $ $ Then $\ m\mid 2n\ $ by  $\,\ 2^m\!-\!1\mid 2^n\!+\!1\mid 2^{2n}\!-\!1.\  $ Note that $\, \ \color{#c00}2 = {2^n\!+\!1-(2^n\!-\!1)}\ $ so   
$2\nmid m\ \ \,\Rightarrow\,\ \  m\mid n \ \,\Rightarrow\,\ 2^m\!-\!1\mid 2^n\!+\!1,\,2^n\!-\!1 \ \Rightarrow\ 2^m\!-\!1\mid\color{#c00}2  \,\Rightarrow\, m=1,\ $ else   
$2\mid m\,\Rightarrow\, m/2\mid n \Rightarrow\!\!\!\!\!\! \underbrace{2^{m/2}\!\!-\!1\mid 2^n\!+\!1}_{\Large 2^{\Large m/2}-1\mid 2^{\Large m}-1\mid 2^{\Large n}+1 }\!\!\!\!\!\!\!, 2^n\!-\!1 \Rightarrow 2^{m/2}\!-\!1\mid\color{#c00}2 \,\Rightarrow\, m=2,\: $ contra $\ m>2$. 
