Let $n = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}, a_i \ge 0$ be the prime factorization of the number $n$. Then the number of divisors of $n$ is $(a_1+1)(a_2+1)\cdots(a_r+1)$.

Let $k$ be a positive integer and let $f(2^k) = p_1^{a_1} p_2^{a_2} \cdots p_r^{a_r}$. Then $a_i = 2^{b_i} - 1$ for some integers $b_i \ge 0$, $i = 1, 2, \dots, r$ and $\sum_{i=1}^r b_i = k$.
Let $p_i, p_j$ be any two distinct prime factors of $f(2^k)$. If we replace $p_i^{2^{b_i}-1}$ with $p_i^{2^{b_i+1}-1}$ and $p_j^{2^{b_j}-1}$ with $p_j^{2^{b_j-1}-1}$ in the prime factorization of $f(2^k)$ we get a number that also has $2^k$ divisors (as $f(2^k)$) because $2^{b_i+1} \cdot 2^{b_j-1} = 2^{b_i} \cdot 2^{b_j}$. Minimality of $f(2^k)$ implies
$$p_i^{2^{b_i}-1} \cdot p_j^{2^{b_j}-1} < p_i^{2^{b_i+1}-1} \cdot p_j^{2^{b_j-1}-1},$$
or equivalently
$$p_j^{2^{b_j}-1} < p_i^{2^{b_i}}. \qquad (\bullet)$$
If we allow some of the integers $c_i$ or $b_i$ to be zero, we can write $f(2^{k+1}) = p_1^{2^{c_1}-1} p_2^{2^{c_2}-1} \cdots p_r^{2^{c_r}-1}$.
Since $\sum_{i=1}^r c_i = k+1 > k = \sum_{i=1}^r b_i$, there is a positive integer $s$ such that $c_s > b_s$.
Let $t$ be a positive integer different than $s$. Using $(\bullet)$ for the pair $(p_s, p_t)$, and for the pair $(p_t, p_s)$ gives
$$p_t^{2^{c_t}} > p_s^{2^{c_s}-1} \ge p_s^{2^{b_s}} > p_t^{2^{b_t}-1}.$$
We conclude $c_t > b_t - 1$, i.e. $c_t \ge b_t$ for all $t = 1, \dots, r$, which implies that $f(2^{k+1})$ divides $f(2^k)$.