Uniquely identifying bits
Your problem is NP-hard. To see this, you can reduce it from the minimum test cover (or test collection) problem: given a set $X$ of $\ell$ elements and a collection $C = \{C_1, \dots, C_k \}$ of $k$ subsets of $X$, find a minimum test cover for $X$ and $C$, i.e., a subset $C'$ of $C$ that has minimum size and satisfies the following property: for every pair of distinct elements $i,j \in X$ there is a set in $C'$ that contains exactly one of $i$ and $j$.

The reduction is as follows: for each element $x \in X$ create a number of $k$ bits $b_1,\dots,b_k$ where $b_i = 1$ if and only if $x \in C_i$.
There is a test-cover $C' = \{ C_{i_1}, C_{i_2}, \dots \}$ if and only if the $|C'|$ bits in positions $i_1, i_2, \dots$ suffice to distinguish all input numbers.

Since the above reduction sets $m=\ell$, the measures for two corresponding solutions of test-cover and your problem coincide, and test cover is not approximable within a factor of $o(\log \ell)$, this also shows that your problem is $o(\log m)$-inapproximable.

On the bright side, the converse reduction works too: given an instance of your problem you can create an instance of minimum test cover by considering the set $X$ containing your $\ell = m$ input numbers $x_1, \dots, x_m$ and $C = \{C_1, \dots, C_k \}$ as a collection of $k = n$ sets where $C_i = \{ x \in X \mid \text{the $i$-th bit of $x$ is } 1 \}$. This means that the $O(\log \ell)$-approximation algorithm for minimum test-cover immediately translates to a $O(\log m)$-approximation algorithm for your problem.

See this paper and this paper.