With Euler's formula $V-E+F=2$ and $E \leq \frac{V(V-1)}2$ (because there cannot be more edges than pairs of vertices), this bound can be obtained: $F \leq \frac{(V-1)(V-2)}2+2$. But I don't know any infinite set of polyhedra with unbounded $V$ for which $F$ would grow faster than linearly. How fast a growth is possible?
I know of Mathematics
Upper bound on number of triangular faces of a convex polyhedron?, even commented there. The difference is I require neither the faces to be triangles, nor the polyhedron to be convex (well, Eulerian, to be able to use his formula in the original form, but even if it isn't, the equality turns into inequality which just allows for the same number or less faces).
Best Answer
[being late, thence just converting the already given comments into an answer, as was required]
Every edge appears on exactly two faces, and every face has at least 3 edges, so we also have $2E≥3F$ (with equality iff all the faces are triangles). Therefore
$$V+F=2+E≥2+\frac32 F$$
This yields the upper bound $F≤2V−4$.
[and moreover]
This is trivially an upper bound, because in the general case [i.e. not using trigonal faces only] we can subdivide any face into triangles without adding any vertices.