Let the greater number be $x$ and the smaller number be $y$.

According to the question:

$\begin{array}{l}
x^{2}-y^{2}=45 \\
y^{2}=4 x
\end{array}$

From (i) and (ii), we get:

$x^{2}-4 x=45$
$\Rightarrow x^{2}-4 x-45=0$
$\Rightarrow x^{2}-(9-5) x-45=0$
$\Rightarrow x^{2}-9 x+5 x-45=0$
$\Rightarrow x(x-9)+5(x-9)=0$
$\Rightarrow(x-9)(x+5)=0$
$\Rightarrow x-9=0$ or $x+5=0$
$\Rightarrow x=9$ or $x=-5$
$\Rightarrow x=9 \quad(\because x$ is a natural number)

Putting the value of $x$ in equation (ii), we get:

$\begin{array}{l}
y^{2}=4 \times 9 \\
\Rightarrow y^{2}=36 \\
\Rightarrow y=6
\end{array}$

Hence, the two numbers are 9 and 6 .