Example 7 Find a rational number $x$, such that $1+5 \times 2^{x}$ is the square of a rational number.


【Analysis】Obviously, when $x$ is a non-integer rational number, $1+5 \times 2^{x}$ is not a rational number, and thus does not satisfy the condition.
It is also obvious that $x=0,1,2$ do not satisfy the condition.
Let's start with the simpler cases.
When $x$ is a positive integer, $x \geqslant 3$.
Assume $1+5 \times 2^{x}$ is the square of a positive odd number $2 k+1$, i.e., $1+5 \times 2^{x}=(2 k+1)^{2}$,
$5 \times 2^{x-2}=k(k+1)$.
Since $(k, k+1)=1$, it is only possible that $k+1=5$, i.e., $x=4,1+5 \times 2^{4}=81$.
If $x=-y$ is a negative integer, then
$$
1+5 \times 2^{x}=\frac{2^{y}+5}{2^{y}}
$$

is the square of a rational number.
Clearly, $y=1$ is not a solution, but $y=2$ is a solution.
Assume $y \geqslant 3, y$ must be even, set $y=2 b$, let $p=2^{b}$. Then
$$
\begin{array}{l}
5+2^{y}=5+p^{2}=q^{2} \\
\Rightarrow 5=(q+p)(q-p) \\
\Rightarrow p=2, q=3 \\
\Rightarrow b=1 .
\end{array}
$$

In summary, the $x$ that satisfies the condition is -2 or 4.