Are there finitely many pairs of integers that are divisible by the same primes such that adding $1$ also keeps them divisible by the same primes?

Question

Integers m and n have the same prime divisors but m is not equal to n, i.e. the same primes are just raised by different powers, resulting in integers m and n. But we also know that m+1 and n+1 have also the same prime divisors.

Are there finitely or infinitely many such integer pairs (m,n)?

Answer 1

There are infinitely many pairs. Any pair of the form $(2^n - 2, 2^n(2^n - 2))$ works for $n$ at least $2$, since $2^n(2^n - 2) + 1 = (2^n - 1)^2$.