Example 9 Find all real numbers $x$ such that $4 x^{5}-7$ and $4 x^{13}-7$ are both perfect squares. ${ }^{[6]}$
(2008, German Mathematical Olympiad)
【Analysis】Let
$$
4 x^{5}-7=a^{2}, 4 x^{13}-7=b^{2}(a, b \in \mathbf{N}) \text {. }
$$

Then $x^{5}=\frac{a^{2}+7}{4}>1$ is a positive rational number, and $x^{13}=\frac{b^{2}+7}{4}$ is a positive rational number.
Therefore, $x=\frac{\left(x^{5}\right)^{8}}{\left(x^{13}\right)^{3}}$ is a positive rational number.
Let $x=\frac{p}{q}\left((p, q)=1, p, q \in \mathbf{Z}_{+}\right)$.
Then $\left(\frac{p}{q}\right)^{5}=\frac{a^{2}+7}{4}$ can only have $q=1$, i.e., $x$ is a positive:

integer.
Obviously, $x \geqslant 2$.
When $x=2$,
$$
\begin{array}{l}
4 x^{5}-7=121=11^{2}, \\
4 x^{13}-7=32761=181^{2}
\end{array}
$$

satisfies the conditions.
When $x$ is an odd number,
$$
a^{2}=4 x^{5}-7 \equiv 5(\bmod 8) \text {, }
$$

which is not valid.
Assume $x$ is a positive even number below.
When $x \geqslant 4$,
$$
\begin{array}{l}
(a b)^{2}=\left(4 x^{5}-7\right)\left(4 x^{13}-7\right) \\
=16 x^{18}-28 x^{13}-28 x^{7}+49 .
\end{array}
$$

Notice that,
$$
\begin{array}{l}
16 x^{18}-28 x^{13}-28 x^{7}+49 \\
\left(4 x^{9}-\frac{7}{2} x^{4}-1\right)^{2},
\end{array}
$$

thus $x \geqslant 4$ and being even is not valid.
In summary, only $x=2$ satisfies the conditions of the problem.