Problem 1533

Prove that for a fixed $m$, assuming $m$ is odd. We will prove that for any positive integer $n$, there exists $k_{n}$ such that $2^{k_{n}}-m$ has at least $n$ distinct prime factors.
(1)
We will prove (1) using mathematical induction:
(1) When $n=1$, $2^{3m}-m$ clearly has at least one prime factor;
(2) Assume $2^{k_{n}}-m$ has at least $n$ distinct prime factors, let $A_{n}=2^{k_{m}}-m$, then $\left(A_{n}, 2\right)=1$, and $2^{k_{n}+\varphi\left(A_{n}^{2}\right)}-m \equiv 2^{k_{n}}-m \equiv A_{n}\left(\bmod A_{n}^{2}\right)$, so $A_{n} \mid 2^{k_{n}+\varphi\left(A_{n}^{2}\right)}-m$,
Take a prime $p \left\lvert\, \frac{2^{k_{n}+\varphi\left(A_{n}^{2}\right)}-m}{A_{n}}\right.$, by $\frac{2^{k_{n}}+\epsilon\left(A_{n}^{2}\right)-m}{A_{n}} \equiv 1\left(\bmod A_{n}\right)$ we know $p \nmid A_{n}$.
Therefore, $2^{p\left(A_{n}^{2}\right)+k_{n}}-m$ has at least $n+1$ distinct prime factors.
By mathematical induction, the conclusion holds.