The Four Ways to Arrange Squares in Barnette Graphs Below you see an example of a bicubic graph consisting of faces with degree $4$ and $6$, which makes up the set of graphs of my interest and is a subset of the so called Barnette graphs. 
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Among the examples in DeLaTorre's "Investigation of Barnette's Graphs" the following fall into my category of interest: 8-1, 12-1, 14-1, 16-2, 18-2, 20-3 (shown above), 20-5 and 20-8.
I found four types of arrangements of square:


*

*three pairs of squares (8-1, 12-1, 18-2, 20-3)

*two triples arranged in row (8-1, 12-1, 16-2, 20-5)

*two triples arranged like a triangle (8-1, 14-1, 20-8) 

*six single squares like in the Truncated octahedron, 24 vertices


The graphs 8-1 and 12-1 are special, since they fit into more than one class. 

Are these four square arrangements the only ones and if so how to prove that?
  Can they be combined, e.g. a single square, a pair and and triple separated by hexagons?

 A: No, they are not the only arrangements: You can also have two pairs of squares along with two isolated squares, like so.

You can also have one pair of squares with four isolated squares, like so (squares are marked "s".)

However, these are the only extra possibilities.
If any three squares meet at a vertex, then the other three squares also meet at a vertex. Because, if you draw three squares meeting at a vertex, and a square adjacent to any of those three, you will find that only two vertices remain not already incident to three edges. Putting an edge between these two results in the graph of a cube. Extending separate edges forces two hexagons around your figure, forming a digon, which is forbidden:

If instead there are only hexagons around your three squares, the outer vertices form the same pattern (3-valent and two-valent alternating), so that adding a square has the same result. Thus there are arbitrarily many rings of hexagons, and finally three squares meeting at a vertex.
Otherwise, if at most two squares meet at any vertex and four or more squares appear in a row, then in fact six squares appear in a cycle and you have a hexagonal prism (Graph 12-1 in the paper).
Otherwise, if three squares appear in a row (but not four), then the other three squares also appear in a row, for similar reasons as the argument for three squares at a vertex: you can have arbitrarily many rings of hexagons around your triple, but the pattern of the outer vertices remains unchanged and finally forces three squares in a row.
