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.
 A: Just some initial observations:

Suppose  $m=2^n-1$ and suppose $m$ is Carmichael. If $p$ is prime and $p \mid m$, then $p -1\mid m-1=2(2^{n-1}-1)$. Since $2^{n-1}-1$ is odd, we must have $p \equiv 3 \pmod 4$ for all $p \mid m$.
For $n \ge 2$, $m \equiv 3 \pmod 4$. $m$ is Carmichael and hence square free. If  $$m = \prod_{i=1}^kp_i\qquad\text{for $p_i$ distinct primes}$$
then
$$\begin{align}
m&\equiv \prod_{i=1}^k3 \pmod 4\\
&\equiv \prod_{i=1}^k(-1) \pmod 4\\
&\equiv (-1)^k \pmod 4
\end{align}$$
So $k$ must be odd, and $m$ is the product of an odd number of distinct primes with $p_i \equiv 3 \pmod 4$.

Certainly $2^n \equiv 1 \pmod m$, and since $2^k<m$ for $m<n$, $2$ has order $n$ modulo $m$. But $$2^{m-1} \equiv 1 \pmod m$$
so $n \mid m-1$. In particular, $2^n \equiv 2 \pmod n$, so $n$ is either prime, a pseudoprime to the base $2$ or $n$ is even and $2^{n-1} = 1 \pmod{\frac n2}$.

None of these conclusions are that restrictive, since we know that a Mersenne number can be prime! I'll try to post more as I think of it.
A: 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$$-1$} .Then $2^t-1= \prod_{i=1}^n{(2^{k_i}+4*m_i-1)} $, from there we understand $n\pmod2=1$ because $(-1)^n==-1$ should satisfy for equality
else $2^{t-1}= even+even+even+...+even+1$ will broke equality.That is where i am at when i do some more i send progress.
A: Your argument must be corrected first. You assume that 2^p-1 nay have the possibility of being Carmichael  number. In the next step you use the theorem of  A  Korsel t (1899) that 2^p-1=m  is a Carmichael  number if and only if m is square free and  every prime factor q  of m is such that q-1|m-1
In  your proof  you have assume that2^p-1=2kp+1=m  which is true.m-1=2^p-2  and it is true that p is a factor of m-1 which is obvious. Also ,k  is a factor of m-1=2^p-2  .
You have not use the theorem correctly .First of all you must find the prime factors of m but not m-1.
Theorem (A. Korselt 1899): A positive composite integer  is a Carmichael number if and only if  is square-free, and for all prime divisors  of  , it is true that  .
You want to know what is k.
2^p-1=(1+1)^p-1=1+C_1^p+-------------+C_(p-1)^p=1+kp
Every  binomial constant is divisible by p for a prime p, then k is obvious. Use your argument in the following way.
Example. Take p=11, 2^p-1=2047=m Prime factors of m are 23 and 89.  m is square free. Take p=23,p-1=22,m-1=2046,22|2046 Now take p=89,p-1=88. Note that 88 does not divide 2046. Therefore 2^p-1 is not a Carmichael number when p=11.
A: Partial Proof
I have found a proof for numbers with an even number of factors, and a prime exponent. ($m = 2^p-1$ and $p > 2$)
First note that $2^p-1 \equiv 3 \mod 4$ for $p > 1$. Next note the following table for $a*b \mod 4$:
$$
\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.
