For every odd prime number $p \gt 3$ there exists another prime number $q \lt p$ such that $p - q = 2^n$ for some $n \geq 1$. Can you prove it?

Question

Conjecture. Let $p$ be an odd prime number greater than $3$. Then there exists another odd prime number $q \lt p$ such that $p - q = 2^n$ for some positive exponent $n$.

Can we prove this or is it another one of those simple to state, but almost impossible to prove, statements about prime numbers?

Answer 1

This seems to be false, with $127$ being the least counterexample.