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.

Partial Proof
I have found a proof for numbers with an even number of factors, and a prime exponent. ($m = 2^p-1$ and $p > 2$)
First note that $2^p-1 \equiv 3 \mod 4$ for $p > 1$. Next note the following table for $a*b \mod 4$:
$$
\begin{array}{c|lcr}
b & a = 0 & a = 1 & a = 2 & a = 3 \\
\hline
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 2 & 3 \\
2 & 0 & 2 & 0 & 2 \\
3 & 0 & 3 & 2 & 1 \\
\end{array}
$$
Because the only results $3 \mod 4$ in the table involve $3 \mod 4$ as a factor. At least one factor of $2^n-1$ must be congruent to $3 \mod 4$ because $2^n-1 \equiv 3 \mod 4$. And because $3*3 \equiv 1 \mod 4$ The amount of such factors must be odd.
Note that factors of $2^p-1$ must be of the form $2kp+1$ for some value $k$. Assuming $p$ to be odd, $p = 2a+1$, and then $2kp+1 = 2k(2a+1)+1 = 4ka+2k+1$. Next $4ka+2k+1 \equiv 2k+1 \mod 4$
Based of the fact that the amount of factors congruent to $3 \mod 4$ have to be odd, if the number of factors of $m = 2^p-1$ is even, then for at least one factor $q$, $q \equiv 1 \mod 4$. Thus $2kp+1 \equiv 1 \mod 4$, so $2k+1 \equiv 1 \mod 4$ and $2k \equiv 0 \mod 4$. Which means $k$ is even.
From a portion proved above, if there is a value $k$ in which $q = 2kp + 1$ for which $q$ is a factor of $2^p-1$, in which $k$ does not divide $2^{p-1}-1$, then $2^p-1$ is never a Carmichael number. Because an even number never divides an odd number, an even $k$ does not divide $2^{p-1}$. This even $k$ value is the one proven above. 
So if the number of factors of $2^p-1$ is even, $2^p-1$ is never a Carmichael number.
