Combinatorics on divisors

I'm not sure if you'll think this is elegant but here goes:-

For convenience, temporarily drop the restriction that $H$ must contain $2$ elements. We can form a variant on Pascal's triangle as follows.

Write $1$ in the top right square i.e. $(4,4)$. This represents the number of sets $H$ with $(4,4)$ as the 'least' point.

Now consider the squares in the next diagonal down i.e. $(3,4)$ and $(4,3)$. For each of these squares we calculate $1$ more than the sum of the numbers above and to the right of it i.e. a total of $2$. For $(3,4)$, the first $1$ represents the number of sets $H$ which only contain $(3,4)$. The sum of the other numbers adds up all those sets $H$ containing $(3,4)$ and at least one other point.

Subsequent diagonals contain the numbers $$4,6,4$$ $$8,16,16,8$$ $$16,40,52,40,16$$ $$96,152,152,96$$ $$416,504,416$$ $$1536,1536$$ $$5136$$

So, I obtain the number $5136$ in square $(0,0)$ i.e. as the number of sets $H$ with $(0,0)$ as the 'least' point.

That means that the sum of all the numbers in the $5\times 5$ block is $ 5136+5135=10271$. We must now exclude the $5^2$ cases where we have only $1$ element in $H$ giving a total of $10246$.