I finally got around to an answer to my question. Let us start with the following observation:
Triangles come in pairs and by looking at these pairs differently, we get a different abstract polyhedron (which is in some sense dual). However, both abstract polyhedron have the same metric and according to the Alexandrov's uniqueness theorem, at most one can be geometrically realizable. Now, it's not too difficult to see that by flipping each pair of triangles likes this, we get that $P_m$ (for $m \geq 6$) is transformed to the skeleton of a gyroelongated $\frac{m}{2}$-gonal bipyramid.
Now we prove that any gyroelongated $n$-gonal bipyramid with congruent isosceles triangular faces is always geometrically realizable for $n > 3$. Such a gyroelongated bipyramid has $n$ rotational symmetry. It consists of two apex vertices and two rings of 'middle' vertices. These two rings form an antiprism. Basically, the figure is a an antiprism with two pyramids pasted on the regular $n$-gons of the antiprism.
By scaling, we can always have that the vertices on the two rings are exactly distance $1$ away from the rotational symmetry axis of the figure.
We need to check that this polyhedron does not have any dihedral angles larger than $\pi$. The dihedral angles between triangular faces of the antiprism are always less than $\pi$, and so are the dihedral angles between the triangular faces of the pyramid. The dihedral angle we need to check is the one between the antiprism and the pyramid (the red edge in the picture above). Note that the purple edges and the symmetry axis lie in a single plane. Therefore, we have the following situation (line through $A$ and $E$ is symmetry axis):
Now, we have the following
- if $\varphi > \psi$, then we have a convex dihedral angle.
- if $\varphi = \psi$, then we have a dihedral angle of $\pi$
- if $\varphi < \psi$, then we have a concave dihedral angle
We know that $\varphi$ and $\psi$ are in between $0$ and $\frac{\pi}{2}$. Therefore, $\varphi > \psi$ iff $\cos(\varphi) < \cos(\psi)$.
Now $\cos(\varphi) = \frac{1-\cos\left(\frac{\pi}{n}\right)}{x}$ and $\cos(\psi) = \frac{\cos\left(\frac{\pi}{n}\right)}{x}$
So $\varphi > \psi$ iff $1-\cos\left(\frac{\pi}{n}\right) < \cos\left(\frac{\pi}{n}\right)$ iff $\cos\left(\frac{\pi}{n}\right) > \frac{1}{2} = \cos\left(\frac{\pi}{3}\right)$.
Which is equivalent to $\frac{\pi}{n} < \frac{\pi}{3}$ or $n > 3$. Therefore, the gyroelongated $n$-gonal bipyramid is always realizable for $n > 3$ independent on the angles of the isosceles faces.
Therefore by the Alexandrov's uniqueness theorem, $P_m$ is not geometrically realizable for $m > 6$. Note that $\varphi = \psi$ iff $n=3$ and thus a gyroelongated $3$-gonal bipyramid is actually a parallelepiped, hence $P_6$ is not geometrically realizable.
The only figure that is left is $P_4$. We prove that it is always geometrically realizable.
Take the following doubly covered polygon where only the red edges are not 'glued'.
If you would make a paper model of this, you would get a 'hole' with four border edges at the top. It is not difficult to see that the left part of the figure has a single degree of freedom (it is a single vertex of degree 4). We can make it so that $F$ moves out of the plane, vertex $A$ and $C$ move to the right and vertex $E$ stays stationary. Since it is symmetric, the right side does not restrict this movement. Then by an argument akin to the intermediate value theorem, there is a point where $A$ and $B$ coincide. By a symmetry argument, it lies in the plane through the point $E$, $F$ and $F$ of the other side.
Note that $C$ and $D$ do not lie in this plane.
Therefore, it forms a polyhedron with the same abstract structure as $P_4$.
The dihedral angles at $AC$, $BD$, $CE$, and $DE$ start at $0$ and increase because the point $F$ moves out of the plane.
They cannot become more than or equal to $\pi$ because that would mean that $C$ and $D$ would lie in the plane of $E$, $F$ and the other $F$.
The dihedral angles at $CF$, $DF$, and $EF$ start at $\pi$ and decrease due to the same reason. Similarly, they cannot become $0$.
The same argument also shows that the dihedral angle of the red lines is strictly convex.
So this shows that $P_4$ is realizable as long as $\alpha^2\beta^2$ and $\beta^4$ are strictly convex vertex figures (even for non-obtuse triangles). We have that $\alpha^2\beta^2$ is strictly convex when $2\pi - 2\alpha - 2\beta > 0$ iff $\pi - \alpha > 0$, which is always the case.
Similarly, $\beta^4$ is always strictly convex because $2\pi - 4\beta > 0$ since $\beta < \frac{\pi}{2}$.
So for every isosceles triangle, $P_4$ is geometrically realizable as a strictly convex polyhedron.
Now onto other classes of polyhedra with congruent isosceles triangular faces!
Best Answer
Unfortunately, you’re right, the proof is just wrong. You don’t even need a weird convex polyhedron; the inequality
$$ V\gt\sum(n-2)f_n $$
is violated for the dodecahedron,
$$ V = 20 \lt 36 = (5-2)\cdot12\;, $$
and for the icosahedron,
$$ V = 12 \lt 20 = (3-2)\cdot20\;. $$
If I had to guess how a correct proof might work, I’d suspect one might be able to use duality (since $v_3+f_3\gt0$ means that either the polyhedron or its dual has at least one triangle). If I find a proof I’ll update the answer.
Update:
Indeed, duality allows for a nice straightforward symmetric proof. If $v_3=f_3=0$, then $V\gt\frac12\sum(n-2)f_n\ge\frac12\sum(4-2)f_n=F$, and by duality analogously $F\gt V$, a contradiction.