Call an integer square-in if it is not square-free or a square. Can two consecutive square-in numbers have a gap of $<8$ integers between them, exactly one of these integers in this 'gap' being a square? If this is possible are there infinitely many?
The equations $x^2-Ny^2=1$ for fixed square-free $N$ have infinitely many solutions; take $N=2$. You can find $3^2- 2. 2^2=1$. Write this as $(3+2\surd2)(3-2\surd2)=1$. Now raising both sides to any $n$ th power you get infinitely many square-in pairs differing by 1. (This problem has been addressed by Brahmagupta several centuries ago and was wrongly called Pell's equation by Euler).
There can only be a maximum of 3 square-free numbers consecutively together in an interval. If two consecutive square-in integers have two consecutive squares between them , say $m^2$ and $(m+1)^2$ with m >=2 then between the two squares there would be >= 4 consecutive square-free integers in this gap; contradiction. So any two consecutive square-in integers could have at most one square between them. So a gap of consecutive square-ins would have a maximum of 8 integers between them.