jmc

Factoring, we have $(x+y)(x-y)=51$. Since $x,y$ are positive, we have $x+y>x-y>0$. Note that $51=51\ast 1=17\ast 3$. Thus either $x+y=51$, $x-y=1$ or $x+y=17$, $x-y=3$. Solving in the first case gives $x=26,y=25$, and the second case gives $x=10,y=7$. Thus there are $\boxed{2}$ pairs $(x,y)$ that solve the equation.