No such polyhedron can have all its faces be regular. One way to see this is to look at the list of Johnson Solids and note that none of them match your description.
You might describe this polyhedron as a near-miss Johnson Solid, though it doesn't seem to be on Wikipedia's list or this one.
Edit: Here's a proof that if the hexagons and pentagons are regular, then the square faces aren't even faces - the vertices won't be coplanar.
First, observe that the truncated octahedron exists, which is effectively this solid with the bottom pentagons converted into squares. Fix two adjacent hexagons near the bottom square, and consider what happens when we widen the angle between them from $90^\circ$ to $108^\circ$ - after some thought, it should become clear that their shared vertex moves "inward" relative to the adjacent two vertices lying on the "equator" of the polyhedron. As this happens (along with the vertically reflected pair of hexagons above), the four points will move out of alignment with each other, and so the face will no longer be any kind of quadrilateral.
(The intuitive idea here is that at most one of a triangular, square, pentagonal, etc., base will be "lucky" enough for the gaps to be nicely coplanar, and the square has already filled that role, so a pentagonally-symmetric solid can't work.)
Edit 2: Here's another proof that the faces can't all be regular.
Suppose otherwise. Let's glue pyramids onto the squares so that the hexagons become extended into trapezoids; the angles of these trapezoids will be $60^\circ$, with the portion lying on the pyramid being an equilateral triangle. If we have a square pyramid with equilateral triangular faces, we're looking at half an octahedron. But then the equator makes a $90^\circ$ turn at every square, so we can't have $5$ of them.