(I suspect a lot of this is known to Seva, but it's probably helpful to have it written down.) 

One lower-bounding argument: Anything that's representable by lines and points is representable by curves and points. When checking if this is so, we can delete any row with just two $1$s, as you can draw a line between any two points. (this generalizes Seva's point in the comments.) Similarly, there is no need for lines to be parallel, so we can delete a column with just two $1$s. Thus we can assume that every row and column has at least three $1$s, which gives a minimum of $7$ vertices: One vertex must be on three edges, each of which contains two other vertices. This is attained uniquely by the Fano plane.

To get an upper-bounding argument, we just need an effective Szemeredi-Trotter theorem. We can do this just by looking at the proof, e.g. on Wikipedia. We construct a graph whose number of edges $e$ is just the number of $1$ entries of the matrix, minus the number of rows. Using the explicit bound $e \leq 4 n$ or $e\leq \left(64 m^2/n^2\right)^{1/3}$. This gives an explicit inequality.

For the projective plane over $\mathbb F_q$, we have $n= q^2+q+1$, $m=q^2+q+1$, $e = q (q^2+q+1)$. Thus it can only be representible for $q \leq 4$. So the projective plane over $\mathbb F_5$ is an example. I guess this isn't very reasonably small.

In fact, I think it's pretty clear that the Fano plane is an example, which must therefore be the smallest. But I don't know how to check this.