Dmitri's answer is definitely correct. I just want to add my geometric intuition, and a generalization, which, in hindsight, is quite obvious.

All in all, we can have the following:

If $P\subset\Bbb R^d$ is a polytope with $n$ facets, each of which is combinatorially (or projectively) equivalent to $Q\subset\smash{\Bbb R^{d-1}}\!$, then for each $k\ge 1$ there also exists a polytope $P_k\subset\Bbb R^d$ with $k(n-2)+2$ facets, all of which are combinatorially (or projectively) equivalent to $Q$.

With this, it should be clear that there are many 4-polytopes with only 5-gonal 2-faces.

The main idea is visualized below.

**Construction:**

- Fix a face $\sigma\subset P$.
- Let $P'$ be the polytope obtained from $P$ by applying a certain projective transformation that a) fixes $\sigma$, and b) moves all vertices of $P$ "beyond" $\sigma$ (see the image). This construction is related to the idea behind the
*Schlegel diagram*, in particular, this transformations always exists.
- Glue $P'$ and $P$ on their common face isomorphic to $\sigma$ (if we have chosen the correct transformation in 2., then this is a convex polytope).

Repeat this to obtain as many $Q$-facets as you like.

Still, it might be interesting to determine the *atomic $Q$-facetted polytopes*, i.e. those, which are not "stacked" in the sense above.

other than" means: not isomorphic as polyhedral complex? (this is a reasonable isomorphism notion; an a priori stronger one would be being isotopic, i.e., have a continuous deformation from one to another) $\endgroup$ – YCor May 23 at 16:38