Bridges across a tiled floor

Let $F(a,b)$ be the number with $a$ specific horizontal bridges and $b$ specific vertical bridges. The other rows and columns may or may not be bridges. Then the number of unaffected squares is $(n-a)(n-b)$ so $$F(a,b)=2^{(n-a)(n-b)}$$
Let $B(n)$ be the number of bridged arrangements.
Now, do inclusion-exclusion:

• Start with single bridges: There are $F(1,0)$ a bridge on the first row, another $F(1,0)$ with a bridge on the second row, and so on, so $nF(1,0)$ in all with a horizontal bridge (counting repetitions). There are $F(0,1)$ with a bridge on the first column, etc, so another $nF(0,1)$ with a vertical bridge.
  $nF(1,0)+nF(0,1)$.
• Two-bridge patterns have been counted twice, so that number must be subtracted. If both bridges are horizontal: there are $n\choose2$ pairs of bridges, each pair has $F(2,0)$ patterns. If both bridges are vertical, there are another ${n\choose2}F(0,2)$ patterns. If one is vertical and the other horizontal, there are $n$ choices for the horizontal one and $n$ choices for the vertical one. In every case, there are $F(1,1)$ patterns.
  Subtract ${n\choose2}F(2,0)+{n\choose1}{n\choose1}F(1,1)+{n\choose2}F(0,2)$
• Three-bridge patterns need to be added back in:
  Add ${n\choose3}F(3,0)+{n\choose2}{n\choose1}F(2,1)+{n\choose1}{n\choose2}F(1,2)+{n\choose3}F(0,3)$
• etc...

The total with no bridges has a slightly simpler formula, and that sum will be $$2^{n^2}-B(n)=\sum_{i=0}^n\sum_{j=0}^n(-1)^{i+j}{n\choose i}{n\choose j}2^{(n-i)(n-j)}$$ The symmetry with $(i,j)$ replaced by $(n-i,n-j)$ gives $$\begin{array}{rcl}2^{n^2}-B(n)&=&\sum_{i=0}^n(-1)^i{n\choose i}\sum_{j=0}^n(-1)^j{n\choose j}2^{ij}\\&=&\sum_{i=0}^n(-1)^i{n\choose i}(1-2^i)^n\end{array}$$ To check for $n=1,2,3$:
$n=1:2-B(1)=0-(-1)=1\to B(1)=1$
$n=2:16-B(2)=1.0^2-2.1^2+1.3^2=7\to B(2)=9$
$n=3:512-B(3)=1.0^3+3.1^3-3.3^3+1.7^3\to B(3)=247$