If both $m$ and $n$ are at least 5, then the number of such matrices is 0. 

Indeed, there must be at three lines containing at least three $1$'s each (up to relabeling the symbols). It is then easy to see that two of those lines must have two $1$'s in the same position (it is impossible to have three 3-subsets of a 5-set such that each pairwise intersection has cardinality at most one.)



Here is a similar argument to show that if $m\geq 3$ and $n\geq 7$, then the number of such matrices if 0.
Considering only the first two rows, we get seven ordered pairs of $\{0,1\}$. We are only allowed to have $(0,0)$ and $(1,1)$ appear once each, so that there are at least five other pairs. In particular there are at least three of one of the other two possibilities, say $(1,0)$. Consider only the three columns where we have $(1,0)$ in the first two rows. In at least two of these columns, the third row must repeat itself, and we get a constant 2x2 sub-matrix.

The answer for $m=1$ and $m=2$ is not hard to calculate explicitly, and the argument above reduces the question to a finite number of other cases :
3x3,3x4,3x5,3x6,4x4,4x5,4x6.