7. Let the pair of positive integers $(x, y)$ satisfy $\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=\frac{1}{\sqrt{20}}$. Then $xy$ has $\qquad$ different possible values.
7.2.
$$
\begin{array}{l}
\text { Given } \frac{1}{\sqrt{y}}=\frac{1}{\sqrt{20}}-\frac{1}{\sqrt{x}}=\frac{\sqrt{x}-\sqrt{20}}{\sqrt{20 x}} \\
\Rightarrow \frac{1}{y}=\frac{x+20-4 \sqrt{5 x}}{20 x}
\end{array}
$$
$\Rightarrow \sqrt{5 x}$ is a rational number.
Since $x \in \mathbf{Z}_{+}, \sqrt{5 x} \in \mathbf{Q}$, therefore, $5 x$ is a perfect square and divisible by 5.
Let $5 x=(5 a)^{2}\left(a \in \mathbf{Z}_{+}\right) \Rightarrow x=5 a^{2}$.
Similarly, let $y=5 b^{2}\left(b \in \mathbf{Z}_{+}\right)$.
Substitute into the equation in the problem:
$$
\begin{array}{l}
\frac{1}{a}+\frac{1}{b}=\frac{1}{2} \Rightarrow 2 a+2 b=a b \\
\Rightarrow(a-2)(b-2)=4 \\
\Rightarrow(a, b)=(3,6),(4,4) \text { or }(6,3) \\
\Rightarrow(x, y)=(45,180),(80,80) \text { or }(180,45) .
\end{array}
$$

Therefore, $x y$ has 2 possible values.