This is an attempt to complement Nick Gill's answer. If it is correct (I am not sure), it should be merged into his answer.

**Proposition**: In every hole-free polygon, there exists a corner square that lies in exactly one maximal subrectangle. 

**Proof**: Observe that as one moves clockwise around the perimeter of the polygon, one must go either left or right at various stages. One obtains a sequence: $R,R,L, R, L \cdots$ and it is clear that the number of $R$'s is 4 more than the number of $L$'s. Therefore there must be a pair of two adjacent  $R$'s.

Assume w.l.o.g. that there is such a pair such that the side between the corners faces west. Consider the maximal rectangle that covers the two $R$ corners (red in the pictures below). This rectangle must run into at least one eastern side. There are two cases:

**Case A**: The eastern side is connected to an $R$ corner adjacent to the $RR$ pair at the north or south:

![Case A][1]

In this case, the middle $R$ corner is contained in exactly one maximal rectangle (the red one) and we are done.

**Case B**: The eastern side is between two adjacent $L$ corners:

![Case B][2]

By the construction, it is clear that this $LL$ pair cannot be used in the same way with any other $RR$ pair. 

Now we know that this cannot happen to all consecutive $R$-corners, because the number of $R$-corners exceeds the number of $L$-corners by $4$.

EDIT: I am not sure about the latter counting argument. What if there is a sequence LLL, such that the first LL pair matches an RR pair, and the second LL pair matches a disjoint RR pair?

![RR RR LLL][3]


  [1]: https://i.sstatic.net/9detb.png
  [2]: https://i.sstatic.net/oXxvD.png
  [3]: https://i.sstatic.net/1jNJo.png