$(m,n)=( {2^k} ({2^k} -2) , ({2^k} -2))$

Lepuslapis wrote: An ordered pair of integers $(m,n)$ with $1<m<n$ is said to be a Benelux couple if the following two conditions hold: $m$ has the same prime divisors as $n$, and $m+1$ has the same prime divisors as $n+1$. (a) Find three Benelux couples $(m,n)$ with $m\leqslant 14$. (b) Prove that there are infinitely many Benelux couples (a) $(2,8), (6,48), (14,224)$ (b) $(2^k-2,2^k(2^k-2))$ for $k\geq2$

(1)the only answers are $(2,8)(6,48)(14,224)$ (2)it suffices to consider $m=2^k-2,n=m(m+2)$ Lengunzoon gave us this problem before~