Given a finite set of convex $d$-dimensional polytopes $\mathcal P$, for some $d\ge 2$.
Question: Is it true that there are only finitely many different convex $(d+1)$-dimensional polytopes whose facets are solely (scaled and rotated versions of) polytopes in $\mathcal P$?
Some clarifications
A face of a polytope is the intersection of the polytope with a touching hyperplane, so subdividing facets does not count here.
In general, two $(d+1)$-polytopes shall be considered as different if they differ not just in scale and orientation.
By the usual rigidity arguments, given the shape of facets and their connections, the metric of the polytope is uniquely determined. Hence, if we can built different polytopes, they will be combinatorially different as well.
Example
There are only finitely many polyhedra that can be built from any finite set of regular polygons, but as far as I know, this result is by enumeration (see, e.g. Johnson solids).
Best Answer
I think this is indeed true. This should follow from the rigidity of convex spherical polytopes and a few more lemmas. I'll explain how to deal with cases $d\ge 3$.
Lemma 1, rigidity. Two convex spherical cobminatorially equivalent polytopes with isomertric faces are isometric.
Lemma 2, bounded volume. The boundary of any convex subset of round $\mathbb S^n$ has volume bounded from above by the volume of round $\mathbb S^{n-1}$.
Lemma 3, gap. Suppose we have a spherical convex polytope in $S^n$ of diameter less than $\pi-\varepsilon$ then the volume of the dual polytope is bounded from below by $c(n,\varepsilon)>0$.
Lemma 4, sum of dual solid angles. Let $P$ be a convex $d+1$-polytope. For each of its vertices consider the dual cone. Then the sum of solid angles of such dual cones is equal to the volume of the round $\mathbb S^d$.
Proof of finitness. So, suppose we have a finite number of convex $d$-polytopes $P_1,\ldots P_k$. Denote by $S_1,\ldots, S_m$ the spherical $d-1$ polytopes corresponding to vertices of $P_1,\ldots, P_k$.
Suppose we have a $d+1$-polytope $P$ with faces homothetic to $P_1,\ldots P_k$. Then for each vertex of $P$ we get a convex spherical $d$-polytope whose faces are composed from $S_1,\ldots, S_m$. Note the the number of such spherical polytopes is finite up to isometry, by Lemmas 1 and 2. Moreover, all of them have diameter less than $\pi-\varepsilon$ for some $\varepsilon$ . So by Lemma 3 the volumes of spherical polytopes dual to those (coming from vertices) are at least $c(d,\varepsilon)$. Hence by Lemma 4 the total number of vertices of $P$ is at most $vol(S^d)/c(d,\varepsilon)$. Hence we have only finite number of combinatorial types of $P$.