I have worked out the following, and would appreciate if anyone can verify or suggest improvements.
If the columns should sum up to $k$, then there are $\binom{m}{k}^n$ $m\times n$-matrices with a column sum of $k$.
Let $d_j$ denote the number of matrices of $j$ rows with at least a non-zero entry but column sum equal to $k$ (and $j \geq k$), then the rest of $m-j$ rows are zeroes and can be chosen in $\binom{m}{m-j} $ ways, thus $$\sum_{j=k}^{m} \binom{m}{m-j}d_j = \binom{m}{k}^n$$
Using Binomial inversion formula we find that $$d_m = \sum_{j=k}^{m} (-1)^{m-j} \binom{m}{j} \binom{j}{k}^n $$
Reversing the order of summation yields $$d_m = \sum_{j=k}^{m} (-1)^{m-j} \binom{m}{j} \binom{j}{k}^n = \sum_{r=0}^{m-k} (-1)^r \binom{m}{m-r} \binom{m-r}{k}^n = \sum_{r=0}^{m-k} (-1)^r \binom{m}{r} \binom{m-r}{k}^n. $$ You can obtain this last summation directly as the inclusion-exclusion formula for $m \times n$ binary matrices with column sum $k$ that avoid the $m$ properties that row $i$ is all zero.