For $n \geq 3$ the gradient of these lines must be non-zero and finite. So it should be possible to give an answer expressed as a sum over all gradients which are reduced fractions $q/p$ with $-n \leq q \leq n$, $\,q \neq 0$ and $1 \leq p \leq n$.
For a given gradient $q/p$ you could count the number points $(x,y) \in \{1,2,\ldots,n\}^2$ that satisfy:
- $1 \leq x+q \leq n$,
- $y+p \leq n$,
- $x-q \geq n+1$ or $x-q \leq 0$ or $y-q \leq 0$,
- $x+2q \geq n+1$ or $x+2q \leq 0$ or $y+2p \geq n+1$.
The first two dot-points ensure that there is a second point $(x+q,y+p)$ on the line. The second two dot-points ensure that there is not too many points on the line -- that is, it ensures $(x+2q,y+2p)$ and $(x-q,y-q)$ (x-q,y-p)$ are both not in $\{1,2,\ldots,n\}^2$.
To give it as a formula, the number of lines that contain exactly two points in $\{1,2,\ldots,n\}^2$ is \[\sum_{-n \leq q \leq n} \sum_{1 \leq p \leq n} \chi(q,p) |B_{q,p}|\] for $n \geq 3$, where $\chi(q,p)=1$ if $\gcd(q,p)=1$, and $\chi(q,p)=0$ otherwise, and $B_{q,p}$ is the subset of $\{1,2,\ldots,n\}^2$ for which the above four dot-points are satisfied.
This should have $O(n^4)$ time complexity (which is not great, but it's better than most formulae I typically deal with).
