Pesky prime

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{p}$

$\dfrac{x+y}{xy}=\dfrac{1}{p}$

$xy=p(x+y)$

$xy=px+py$

$xy-px-py=0$

$xy-px-py+p^2=p^2$

$(x-p)(y-p)=p^2$

Now in general, $p^n$ has $(n+1)$ distinct positive factors for any prime $p$ and any nonnegative integer $n$ , considering that $p$ is irreducible in the ring of integers. To be exact, these factors are $p^i$ for $i=0,1,...,(n-1),n$ Thus $p^2$ has $2+1=3$ distinct positive factors and without loss of generality, consider $x-p=1, p, p^2$ . Note that all the corresponding negatives, namely $-1, -p, -p^2$ are also factors of $p^2$ but we have to exclude $-p$ as a possible value of $(x-p)$ as this would make $x=0$ .

Thus we have a total of $3\times2-1=6-1=\boxed{5}$ distinct solutions for $(x,y)$ .