Given convex polyhedra $P, Q$, say that one can inscribe $P$ in $Q$ if we can find points on the surface of $Q$ whose convex hull is similar to $P$.
If we restrict $P, Q$ to be Platonic solids, we can achieve every case except inscribing the dodecahedron into the tetrahedron, cube, or octahedron; the other $17$ distinct pairs work.
For the $17$ working cases, there are "nice" constructions, where the solids are positioned in symmetric ways with respect to each other. Consequently, such constructions are pretty easy to verify by a combination of symmetry arguments, direct computation of one or two distances, and arguments from degrees of freedom / continuity. To see examples of these pairs, Moritz Firsching's 2018 paper "Computing maximal copies of polytopes contained in a polytope" (PDF link via arXiv.org) shows instances of maximal containment, most of which have "nice" symmetry (or are obviously given by e.g. taking duals). Perhaps the hardest case to see a "nice" construction for is the cube inside the tetrahedron, pictured below (bold edges tangent to the faces):
For two of the three impossible cases, I have simple proofs of impossibility:
If a dodecahedron were inscribed in a tetrahedron, every face would need to have a pentagon on it, but the dodecahedron doesn't have four mutually disjoint faces.
If a dodecahedron were inscribed in a cube, then any face with at least three of the vertices of the dodecahedron would have to have an entire pentagon on it (and no more of the vertices), since three points determine a plane and the dodecahedron is convex. Then (by simple counting) at least three of the cube's faces must have such a pentagon on them, so two such are orthogonal to each other. But the dodecahedron has no orthogonal faces.
However, I know of no such proof that the dodecahedron cannot be inscribed in the octahedron; it seems fairly plausible, at first glance, that one could mount the dodecahedron with two opposite faces tangent to opposite faces of the octahedron and arrange the other $10$ points across the other $6$ faces of the octahedron somehow. The simplest proof I know involves doing some gross trigonometry involving the necessary side length of such a dodecahedron and the inradius of the regular $10$-gon formed by taking an equatorial cross-section of the dodecahedron in between these two opposite faces.
Is there a conceptually simple proof that this construction is impossible? I realize this is somewhat subjective, but I hope the intended style of argument is clear: reasoning from symmetry and combinatorial incidences, rather than bashing the coordinates of different vertices.