Is there a 4-polytope without 3-gonal and 4-gonal faces, other than the 120-cell? The question is in the title:

Question: Is there any 4-dimensional polytope without 3-gonal and 4-gonal faces (of dimension two), other than the 120-cell?

I consider only convex polytopes (convex hull of finitely many points) that are full-dimensional (not contained in a proper subspace).
And I consider a polytope to be distinct from the 120-cell if it has a non-isomorphic face-lattice.
It is known that any 4-polytope must have a 3-gonal, 4-gonal or 5-gonal face of dimension two.
The 120-cell has only 5-gonal faces of dimension two.
 A: 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.
A: 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.
