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.

EDIT : I've edited the argument to make it stronger Suppose that $m\geq 3$ and $n\geq 5$ so that there is a 3x5 submatrix A. I show that the number of possibilities is zero in this case.

In A, there are at least two rows with at least there $1$'s each (up to relabeling the symbols). Since we cannot have a constant 2x2 submatrix, we may assume that the first two rows of the matrix are [11100] and [00111].

To avoid a 2x2 contant submatrix, the first two entries of the third row must be different, but then, whatever choice we make for the third one, we will get a constant 2x2-submatrix in the first and third row.

The answer for $m=1$ and $m=2$ is not hard to calculate explicitly.

Together with the answer above, this reduces the problem to checking the following cases (which is not too hard): 3x3,3x4,4x4.

This is not a complete answer but 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.)

If one of $m$ or $n$ is at most 2, it is pretty easy to find an explicit number. Maybe it is also possible for 3 and 4.

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.