(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.

I think a simple Euler characteristic bound might work better. If there are $P$ points and $C$ curves, $I$ incidences, and $c$ bonus crossings where two curves cross away from a points, we get a graph in the plane with $P+c$ vertices and $I-C +2c$ edges, so it has $2+I+c-P-C$ faces. With the inequality $3F \leq 2E$, since two curves cannot intersect in more than two points, so all faces are triangles are larger, we get the inequality:

`\[ 6+ 3I + 3c -3P - 3C \leq 2I - 2C+4c\]` 

`\[ I \leq 3P + C + c-6\]`

For the projective plane, $c=0$ because each pair of edges already intersect at a point. $I=(q+1)(q^2+q+1)$, $P=q^2+q+1$, $C=q^2+q+1$, so we get

`\[(q+1)(q^2+q+1) \leq 4 (q^2+q+1) - 6\]`

which improves it to the projective plane over $\mathbb F_3$.