There are other polytopes. To construct one let's do the following. Remember first that in the hyperbolic $4$-space there exists a regular compact *right-angled* 120-cell. Here, right-angled means that any two adjacent faces intersect under angle $\frac{\pi}{2}$. Regular means, that all the faces are isomeric, and the polytope has the same group of self-isometries as the Euclidean 120-cell. This polytope is discussed, for example, in

https://pdfs.semanticscholar.org/a0eb/ccbed0687d966a9aaaac2f370bc930a556be.pdf

at the bottom of page 65. The references to more classical articles are given there.

Now, if we double it in one face then we get a new convex polytope, and it is not hard to see, that it doesn't have 2-faces that are triangles and quadrilaterals. But any convex hyperbolic polytope is also combinatorially equivalent to a Euclidean one.

More generally, you can take any compact right-angled hyperbolic polytope in $\mathbb H^4$. Since it is hyperbolic and right-angled, it can not have $2$-faces that are triangles of quadrilaterals. And there is a infinite number of such polytopes in dimension 4. Each of them gives a Euclidean one as well.

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