TITLE: Prove that the set of integers of the form $2^k−3$ contains an infinite subset of relatively prime numbers
QUESTION [10 upvotes]: Prove that the set of integers of the form $2^k − 3(k = 2, 3, ...) $
  contains an infinite subset in which every two members are relatively
  prime.


Trying to solve this problem, I stumbled on this conjecture: 
There is an infinity of prime numbers $p$ such that $p=2^n - 1, n\in \mathbb{N}$
I'm interested in solving both the problem and the conjecture.

REPLY [2 votes]: Let $p_i$ be the $i^\text{th}$ odd prime. Let $(k_r)_{r=1}^\infty$ be any sequence of integers such that $k_1 \geqslant 3$, $k_{r+1} > k_r$, and $k_{r+1}$ is divisible by $\prod_{i=1}^{m_r} (p_i - 1)$, where $p_{m_r}$ is the largest prime factor of $a_r = 2^{k_r} - 3$. Then
$$
a_{r+1} = \left(2^{k_{r+1}/(p_i - 1)}\right)^{p_i - 1}\!\!\! - 3
\equiv -2 \pmod{p_i} \quad (i = 1, \ldots, m_r),
$$
by Fermat's little theorem. Therefore, $a_{r+1}$ has no prime factor in common with $a_r$; and $m_{r+1} > m_r$. The sequences $(k_r)$, $(m_r)$, $(a_r)$ are strictly increasing. If $r, s$ are integers with $1 \leqslant r < s$, then $a_s$ is not divisible by any of the primes $p_1, \ldots, p_{m_{s-1}}$, in particular $p_1, \ldots, p_{m_r}$, therefore $a_s$ is prime to $a_r$.