Combinatorial proof that 18-faced convex deltahedron does not exist?

I probably don't understand your question, because there are some convex Johnson solids that have 18 faces, such as the "elongated square cupola". See link.

V - E + F = 20 - 36 + 18 = 2

Or the "square orthobicupola", 2nd link.

V - E + F = 16 - 32 + 18 = 2

Can you restate or refine your question? Do you mean, for example, if I were to start with an icosahedron and removed 2 triangular faces and tried to put it back together? Are the faces still supposed to be equilateral triangles? Because if they don't have to be, then it's quite possible to morph a icosahedron into a convex 18 sided polyhedron with only triangular faces. To see how, take a cluster of 5 faces and replace them with 3, and adjust the locations of the vertices so that we have a convex polyhedron with 18 distinct faces. Yes, it would mean that there will be 8 vertices with 5 triangles, 2 with 4, and 1 with 6, but this works only if we used some non-equilateral triangles. This polyhedron would have to have 11 vertices in which the 54 corners of the 18 triangles would have to be bunched up into, and in this case, 8*5 + 2*4 + 1*6 = 54, and 8 + 2 + 1 = 11. I'm sure there are other ways.

Edit: As a matter of fact, I think there is a way to form a 18 sided convex polyhedron using equilateral triangles. If you leave this open, I might be able to work out the graphic on how this is done. It has 11 vertices, so that we have V - E + F = 11 - 27 + 18 = 2, as it should be, and that the breakdown of the vertices is as follows, 8*5 + 1*4 + 2*3 = 54. This is not a sure bet, I have to work out the numbers ensure that this is dimensionally possible.

Edit 2: Okay, see graphics, 2 more links---2 views of the same polyhedron. It is possible to assemble 18 equilateral triangles into a convex polyhedron. At the bottom, it appears to the eye that there are 2 pairs of faces that are coplanar, but there is the barest of outward kink in each pair of faces, so that this is in fact a convex polyhedron. See last link, orthogonal view, and note that tiny outward kink.

There cannot be a convex 18 sided deltahedron! I am working on the proof in a topology class. The shape from the above answer does have coplanar faces. It is obvious from the last picture. It may appear to have an "outward kink" but does not. Observe by adding angle of triangles