Can a Mersenne number ever be a Carmichael number? Can a Mersenne number ever be a Carmichael number?
More specifically, can a composite number $m$ of the form $2^n-1$ ever pass the test: $a^{m-1} \equiv 1 \mod m$ for all intergers $a >1$ (Fermat's Test)?
Cases potentially proved so far: (That are never Carmichael numbers)


*

*where $n$ is odd

*where $n$ is prime


Work using "main" definition:
First off take the definition of a Carmichael number:

A positive composite integer $m$ is a Carmichael number if and only if  $m$
  is square-free, and for all prime divisors $p$ of $m$, it is true that $p -
1 \mid m - 1$.

Let's assume $m=2^n-1$ is squarefree. (Best case, and I believe it always is for $2^p-1$)
Take the case where $n$ (in $2^n-1$) is a prime $p$. All factors of $2^p-1$ must of the form: $2kp+1$ for some constant $k$. So will $2kp$ ever divide $2^p-2$? Factoring a $2$ out gives us $kp \mid 2^{p-1}-1$, or split into two: $k \mid 2^{p-1}-1$ and $p \mid 2^{p-1}-1$ must both be true. By Fermat's little theorem, $2^{p-1} \equiv 1 \mod p$, so $p \mid 2^{p-1}-1$ is always true. 
So if $k \mid 2^{p-1}-1$ for $k = {q-1 \over p}$, is false for at least one factor $q$ of $2^p-1$, no Carmichael numbers can exist of form $2^p-1$. 
Now for other cases where $n$ is composite, lets say $n=cp$, for some prime $p$, and some number $c$:
$\begin{align}2^{cp}-1&=(2^p-1)\cdot \left(1+2^p+2^{2p}+2^{3p}+\cdots+2^{(c-1)p}\right)\end{align}$
Thus: $2^{n-1} \mid 2^p-1$
Because of that, we must look at the factors of $2^p-1$ when considering if $2^{cp}-1$ is a Carmichael number. So we know those factors are already of form $2kp+1$, and then $kp \mid 2^{cp-1}-1$.
This is where I'm left. on an incomplete proof.
Using Bernoulli definition: 

An odd composite squarefree number $m$ is a Carmichael number iff $m$
  divides the denominator of the Bernoulli number $B_{n-1}$.

Using the Von Staudt–Clausen theorem, there may be a way to proof that that factors of the Bernoulli number denominators may never divide a mersenne number.

Let say $2^t-1$ has n prime factors like {$p_1,p_2,p_3,..,p_n$}.
If it is Carmichael number than $2^t-2$ should divisible by all n numbers which are like
{$num_1=p_1-1,num_2=p_2-1,num_3=p_3-1,..,num_n=p_n-1$} and all those numbers are even.
$2^t-2=2*(2^{t-1}-1))$ because of that equality n numbers all should just be just divisible by 2 once.Then the prime factors should like {$p_1=2^{k_1}+4*m_1-1,p_2=2^{k_2}+4*m_2-1,p_3=2^{k_3}+4*m_3-1,..,p_n=2^{k_n}+4*m_n$$-1$} .Then $2^t-1= \prod_{i=1}^n{(2^{k_i}+4*m_i-1)} $, from there we understand $n\pmod2=1$ because $(-1)^n==-1$ should satisfy for equality
else $2^{t-1}= even+even+even+...+even+1$ will broke equality.That is where i am at when i do some more i send progress.
