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.

Your argument must be corrected first. You assume that 2^p-1 nay have the possibility of being Carmichael  number. In the next step you use the theorem of  A  Korsel t (1899) that 2^p-1=m  is a Carmichael  number if and only if m is square free and  every prime factor q  of m is such that q-1|m-1
In  your proof  you have assume that2^p-1=2kp+1=m  which is true.m-1=2^p-2  and it is true that p is a factor of m-1 which is obvious. Also ,k  is a factor of m-1=2^p-2  .
You have not use the theorem correctly .First of all you must find the prime factors of m but not m-1.
Theorem (A. Korselt 1899): A positive composite integer  is a Carmichael number if and only if  is square-free, and for all prime divisors  of  , it is true that  .
You want to know what is k.
2^p-1=(1+1)^p-1=1+C_1^p+-------------+C_(p-1)^p=1+kp
Every  binomial constant is divisible by p for a prime p, then k is obvious. Use your argument in the following way.
Example. Take p=11, 2^p-1=2047=m Prime factors of m are 23 and 89.  m is square free. Take p=23,p-1=22,m-1=2046,22|2046 Now take p=89,p-1=88. Note that 88 does not divide 2046. Therefore 2^p-1 is not a Carmichael number when p=11.
