See arxiv.org/abs/math/9902004 and arxiv.org/abs/math/0503507.

I don't know much about them, but you can look them up. They're also called Thron-type continued fractions. Two references are projecteuclid.org/journals/… and jstor.org/stable/44237585 .

The second fraction is called a T-fraction.

Incidentally, $(2m)!\,(2n)!/m!\,n!\,(m+n)!$ is the constant term in $(1+x)^m(1+1/x)^m(1-x)^n(1-1/x)^n$.

For solving the general quintic, see also math.stackexchange.com/questions/540964/… and en.wikipedia.org/wiki/Quintic_function.

You have to delete a 2-dimensional slice. That's what corresponds to a vertex. So the analogous statement for $n$-dimensional arrays will be about arrays in which every $n-1$-dimensional slice has at least one 1.

By the same reasoning the exponential generating function for (labeled) graphs without isolated vertices is $e^{-x}\sum_{n=0}^\infty 2^{\binom n2} x^n/n!$; these correspond to symmetric 0-1 matrices with 0s on the diagonal and a 1 in every row. (oeis.org/A006129)

No, there's nothing special about two dimensions. The analogous statement will be true for 3D arrays. Think of a 3D 0-1 arrays as tricolored 3-regular hypergraph: you have three vertex sets each a different color, corresponding to the three coordinates, and each 1 of the array corresponds to a triple of vertices, one of each color. Each such hypergraph will be a set of isolated vertices together with a hypergraph without isolated vertices.