Solution below. Perhaps this is also well known. I post it only because it is pretty.

Solution: Let the bases of the pyramids tile the plane with squares, draw a second plane through the apexes of the pyramids, and let a second set of inverted pyramids tile this plane with their bases, the corners of each base coinciding with the apexes of the upright pyramids. Since the volume of a pyramid is 1/3·h·(base area), while the volume of a rectangular slab is h·(base area), the pyramids together fill 2/3 of the volume of the slab. The spaces between the pyramids are rather obviously perfect regular tetrahedrons, and since each of these shares a face with 4 pyramids, while each upright pyramid shares a face with 4 tetrahedrons, they are equinumerous. Hence, since they are equal in number and fill the remaining 1/3 of the volume, their individual volumes must be identical to those of the pyramids.

Let's start with just four 4-pyramids whose bases form a square. The space between any adjacent two can be filled with a 3-pyramid by joining their apexes. These four lines form a base of an inverted 4-pyramid and, if we put one more 4-pyramid on top of that base, we'll get a 4-pyramid twice the size of the original one. This big 4-pyramid consists of 6 small 4-pyramids and four 3-pyramids. If V is the volume of a small 4-pyramid, W that of a 3-pyramid, then

8V = 6V + 4W,

meaning that V = 2W.

Where are the holes (pun intended) in this argument?