A regular tetrahedron can be dissected into $1,2,3,4,6,8,12,$ or $24$ congruent pieces. Is this it? By placing a tetrahedron on a face and making vertical cuts centered at the "top" vertex, it is easy to dissect the tetrahedron into $1, 2, 3,$ or $6$ congruent pieces.
By cutting the tetrahedron into four identical pyramids meeting at the center, one for each face, we can further subdivide these pyramids to yield a total of $4, 8, 12,$ or $24$ congruent pieces.
However, it's not obvious to me how to do anything more interesting than this in a way that yields dissections into other numbers of pieces - unlike the triangle, which can be subdivided into many smaller equilateral triangles, the tetrahedron does not decompose into congruent smaller regular tetrahedra, so there is no natural way to 'bootstrap' these constructions to higher numbers of pieces. In particular, I'm most interested in the question of whether a tetrahedron can be dissected into arbitrarily high numbers of pieces.
Some options that come to mind:

*

*Some of the pieces resulting from one of the above dissections might have a further dissection I haven't thought of.


*Polyforms on the tetrahedral-octahedral honeycomb could potentially work to tile a tetrahedron of large side length, just as some polyominoes tile a square and some polyiamonds tile a larger triangle. (However, since the ratio of tetrahedra to octahedra in a given finite tetrahedral portion of the honeycomb monotonically decreases with the side length of said tetrahedron, any given polyform will work with at most one scale.) See also this question on MSE.


*Perhaps something like Dehn invariants could to used to attempt a proof of impossibility, somehow? I'd be skeptical, though.
 A: At the question tetrahedron into similar tetrahedron are two examples of non-regular tetrahedra that can be divided into 8 copies of themselves. They each have some symmetry, so they are examples of non-regular tetrahedra that can be divided into 16 identical shapes. And they can be extended to more pieces.
Since you're using regular tetrahedra, you won't find a nice orthoscheme to allow that trick.
Not a proof:
The pieces of any shape would need to have portions of the arccos(1/3)~70.5288 degree dihedral angle. To divide up the existing angle symmetrically I believe you're trapped by symmetries of the octahedron and need a subgroup of $O$. To place that angle internally you'd need the hypothetical shape to handle that, which implies portions of an octahedron. If such a single piece existed, it would likely be nice enough to be a space-filler.  If that could be proven, then a look-up could be done in On Space Groups and Dirichlet-Voronoi Stereohedra to see if the shape exists.
A: The easiest approach is probably to consider the symmetries of the spherical tetrahedron as a kaleidoscope. Here, lines of mirror symmetry on the sphere are treated as kaleidoscope mirrors.
The tetrahedral kaleidoscope is illustrated, along with the others, in Rouse Ball and Coxeter; Mathematical Recreations and Essays (13th Edn, Dover, 1987, p.158).
Coxeter devotes a chapter of Regular Polytopes to a fuller discussion of such kaleidoscopes.
Note that in the case of the tetrahedral kaleidoscope, the polyhedron's duality of faces and vertices means that what appear to be 24 isosceles triangles on the sphere are scalene when drawn on the actual tetrahedron and occur in alternating left- and right-handed forms.
These represent the bases of irregular triangular pyramids (i.e. irregular tetrahedra) with their apex at the centre. They are the characteristic tetrahedra or orthoschemes of the regular polyhedron, and cannot in general be further subdivided into congruent parts.
The tetrahedral kaleidoscope yields just the congruent sets which the question enumerates, and no more.
Are there any asymmetric or low-symmetry solutions? For example there is at least one such dissection of the cube into three congruent pieces. It is difficult if not impossible to prove that no such "black swan" solutions exist, but (to the best of my knowledge) none is known for the regular tetrahedron.
