2 for 2

If $X=p_1^{\alpha_1}\dots p_n^{\alpha_n}$ does not have $2$ as one of its primes, then

$f(X)=(\alpha_1+1) \dots (\alpha_n+1) \ , \ f(2X)=2. (\alpha_1+1) \dots (\alpha_n+1)$

the difference would be

$f(2X)-f(X)=(\alpha_1+1) \dots (\alpha_n+1)=2$

As $\alpha_i \geq 0$ there can be only one prime with power $1$ .

If $X=2^{\alpha_1}\dots p_n^{\alpha_n}$ , then

$f(X)=(\alpha_1+1) (\alpha_2+1)\dots (\alpha_n+1) \ , \ f(2X)=(\alpha_1+2) (\alpha_2+1)\dots (\alpha_n+1)$

the difference would be

$f(2X)-f(X)=(\alpha_2+1)\dots (\alpha_n+1)=2$

With the same reasoning as before, there can be only one prime (other than $2$ ) with power $1$ .

So, taking the two cases into account, $X$ can have a maximum of $2$ prime factors.