The required minimum is $n$.
To prove this, notice that the sum of two like powers of 2 is again a power of 2, so the number of powers of 2 that add up to a positive integer $k$ can successively be decreased while keeping the sum constant. The process then ends up with $k$ being expressed as a sum of pairwise distinct powers of 2; that is, the binary expansion of $k$. At this stage, the number of summands can no longer be decreased, so the smallest number of powers of 2 that add up to $k$ is equal to the number of units in the binary expansion of $k$.

1st Proof.
Since multiplication by a power of 2 does not change the number of units in a binary expansion, we may and will assume that $m$ is odd. Leaving aside the trivial case $m = 1$, let $m \ge 3$ so its binary expansion is $m = 1 + \sum_{i=1}^{p} 2^{k_i}$, where $0 < k_1 < k_2 < \dots < k_p$; and since $m < 2^n$, it follows that $k_p < n$, so $p < n$. Now, write $m \cdot (2^n - 1) = 2^n \cdot (m-1) + ((2^n - 1) - (m-1)) = \sum_{i=1}^{p} 2^{n+k_i} + (\sum_{i=0}^{n-1} 2^i - \sum_{i=1}^{p} 2^{k_i})$.

The first sum consists of exactly $p$ pairwise distinct powers of 2, each of which is greater than $2^n$, and the number in parentheses is the sum of exactly $n - p$ pairwise distinct powers of 2, each of which is less than $2^n$.
Consequently, $m \cdot (2^n - 1)$ is the sum of exactly $p + (n - p) = n$ pairwise distinct powers of 2; that is, its binary expansion has exactly $n$ units, as stated.

2nd Proof.
If $m = 2^k$ for some $k < n$, then $m \cdot (2^n - 1) = 2^{n+k-1} + 2^{n+k-2} + \dots + 2^{k+1} + 2^k$, so the binary expansion of $m \cdot (2^n - 1)$ has exactly $n$ units.
If $m$ is not a power of 2, let $m = \sum_{i=1}^{p} 2^{k_i}$, where $p \ge 2$ and $0 \le k_1 < k_2 < \dots < k_p$, be the binary expansion of $m$, to write $m \cdot (2^n - 1) = m \cdot 2^n - m = (m \cdot 2^n - 2^{n+k_1}) + (2^{n+k_1} - m) = \sum_{i=2}^{p} 2^{n+k_i} + (2^{n+k_1} - \sum_{i=1}^{p} 2^{k_i})$.
Clearly, the first sum is an integer greater than $2^{n+k_1}$ whose binary expansion has exactly $p-1$ units.
The number in parentheses is a positive integer ($m < 2^n \le 2^{n+k_1}$) smaller than $2^{n+k_1}$ whose binary expansion has exactly $n+k_1-k_p+\sum_{i=1}^{p-1}(k_{i+1}-k_i-1) = n-p+1$ units.
Consequently, the binary expansion of $m \cdot (2^n - 1)$ has exactly $(p-1) + (n-p+1) = n$ units, as stated.