smc

We can simplify $\sqrt{1984}$ to $8\sqrt{31}$. Therefore, the only solutions are $a\sqrt{31}+b\sqrt{31}$ such that $a+b=8$ and $0<a<b$. The only solutions to this are $a=1,b=7;a=2,b=6;a=3,b=5$. Each of these gives distinct pairs of $(x,y)$, so there are $3$ pairs, $\boxed{3}$.