Stonebridge's method is probably the simplest and most straightforward. He is basically using superposition of potential, along with the known potential of an individual point charge. There is another way to get the answer through symmetry arguments and the definition of potential. Since, I'm assuming you can get the answer very easily with Stonebridge's good suggestion , I'm not afraid of giving the answer away. I mention this alternative way as a learning tool, not because I think it is the better way to solve this particular problem.
Assuming static electric fields, and hence conservative forces, you can make the following argument. Remember the definition of potiential is the net work done (per unit charge) in moving a charge from infinity to any finite point near the charges.
If you travel (with a test charge) on the x-axis from x=-inf to a point left of the left-most charge, it's clear potential can't be zero because the test charge will always experience a net positive (or net negative) force along the entire path. The reason is that the force direction caused by the closer charge always dominates. The same argument can be made for a path along the x-axis going from x=+inf to a point right of the right-most charge. Hence, a zero potential point, if it exists on the x-axis, must be between the charges.
The path you take going to the portion of the x-axis between the charges should not come from infinity on the x-axis, but preferably along a path on the yz-plane from infinity to the origin, and then along the x-axis. Traveling along the x-axis from infinity will bump you into one of the charges which creates a singularity which is better avoided. Also, the benefit of this yz-plane path is that any force from the electric field is always perpendicular to the direction of travel on the path, hence work is zero, and potential stays zero on this entire V=0 equipotential surface.
So it is clear that the entire yz-plane, which includes x=0, is a V=0 surface. This agrees with the methodology offered by Stonebridge because any point in the yz-plane has equal distance to the two charges, hence the equal and opposite potentials cancel. It is also clear that no other point between the charges can have V=0 because a force exists in one direction (parallel to the x-axis) as you move further along the x-axis, from the origin.