I stand corrected.

The answer John is looking for is to use Find and Replace with wildcards

Search for ([\]])([\[]) and replace with \1^p\1.

It looks complicated because both [ and ] have to be escaped by preceding them with the \

The replace box refers to fiels which are defined by everything between 9 and 0. The search has two such fields.

the ^p is the replace special symbol for a pararaph marker/ NOTE that ^p cannot be used in the search box, you need to use ^013 instead.

If you were doing this by a macro then

.find.text="([\]])([\[])"
.replacement.text="\1^p\2"