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If $2^n-1$ is composite then $n$ is composite. Where n is larger than 1

How can I disprove this using a counter example?

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    You need to find a specific $n$ where the hypothesis "$2^n$ is composite" is true, but the conclusion "$n$ is composite" is false. – 2017-01-31
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    Sorry guys I misses a part in the problem but I've uppdaterad it now – 2017-01-31
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    @O.VonSeckendorff: Any prime? How about $n=2$? – 2017-01-31
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    @AsafKaragila The question was edited. $n=2$ would have done the job in the original post – 2017-01-31

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It is well known that $2^n-1$ can only be prime if $n$ is prime (A prime of the form $2^n-1$ is called a Mersenne-prime)

But the converse is not true. There are primes $n$, such that $2^n-1$ is composite, for example $n=23$, refuting the claim. The first few counterexamples are

? forprime(p=1,100,if(isprime(2^p-1)==0,print1(p," ")))
11 23 29 37 41 43 47 53 59 67 71 73 79 83 97
?
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    Great answear! I just have a question, how do I actually find the answear 23 and proove it mathematically, do I have to test every integrerat up to 23? It looks like you are using a programming language – 2017-01-31
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    In fact, for most of the primes $n$ , the number $2^n-1$ is composite. Mersenne-primes are quite rare. It is conjectured that infinite many Mersenne-primes exists. Surprisingly, it is even unknown whether infinite many primes $n$ exist such that $2^n-1$ is composite. But I would be very surprised indeed, if $2^n-1$ would be actually prime for every sufficient large prime $n$ – 2017-01-31
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    @pillikasnazsaxc Yes, I used a program, but Mersenne-numbers are very well known. Just google. – 2017-01-31
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    The equation $2^{11}-1=2047=23\cdot 89$ can be verified by hand, and it is clear that $11$ is prime. This would be a formal disproof (for the purists). But we can even find a (probably) infinite family of counterexamples by proving the following (which can relatively easy be done) : If $p$ is a prime of the form $4k+3$, such that $2p+1$ is prime as well, then $2^p-1$ is divisble by $2p+1$, hence composite. – 2017-01-31
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    The inequality $2^p-1>2p+1$ for $p\ge 4$, can be easily shown by induction to show that $2p+1$ actually is a proper divisor. – 2017-01-31