If an integer, $2p + 1$, where $p$ is a prime number, is a divisor of the Mersenne number $2^p - 1$, then $2p + 1$ is a prime number.
My argument is that because divisors of the Mersenne number $2^p - 1$ can’t be $< p$ if $p$ is a prime number. Therefore if $2p +1$ is a divisor of $2^p - 1$ it has no divisors as $p$ is $>$ the square root of $2p + 1$. This will therefore make $2p + 1$ a prime number.
Is this proof correct?
Your idea is good, but you should prove that every divisor must exceed $p$.
In general, we can prove that for a prime number $p$ and an odd prime $q$ dividing $2^p-1$ , we must have $p|q-1$
The proof is as follows : Let $k$ be the smallest positive integer with $2^k\equiv 1\mod q$. Because of $2^p\equiv 1\mod q$, $k$ must be a divisor of $p$.
$k=1$ is impossible because this would imply $q|1$. Hence , $k=p$
Because of $2^{q-1}\equiv 1\mod q$ (Fermat's little theorem) we can conclude $p|q-1$
In particular, $2p+1$ , if it divides $2^p-1$ must be prime because every prime factor must be of the form $2kp+1$ with positive integer $k$ , hence there cannot be a prime factor less than $2p+1$