Example 1 Find all integers $x$ such that $1+5 \times 2^{x}$ is the square of a rational number. ${ }^{[1]}$
(2008, Croatia National Training (Grade 2))
Discuss the following two scenarios:
(1) If $1+5 \times 2^{x}$ is a perfect square, then $x \in \mathbf{N}$. Let $1+5 \times 2^{x}=y^{2}(y \in \mathbf{N})$. Hence, $(y+1)(y-1)=5 \times 2^{x}$.
If $x=0$, then $y^{2}=6$, which is impossible. Therefore, $x \neq 0$.
Since $y+1$ and $y-1$ have the same parity, both must be even.
(i) If $\left\{\begin{array}{l}y+1=2^{\alpha}, \\ y-1=5 \times 2^{\beta},\end{array}\right.$ where $\alpha, \beta \in \mathbf{N}_{+}$, $\alpha+\beta=x$ and $\alpha>\beta$.
Subtracting the two equations gives $2^{\beta}\left(2^{\alpha-\beta}-5\right)=2$.
Thus, the odd number $2^{\alpha-\beta}-5=1$, i.e., $2^{\alpha-\beta}=6$.
This is impossible.
(ii) If $\left\{\begin{array}{l}y+1=5 \times 2^{\alpha} \\ y-1=2^{\beta},\end{array}\right.$ where $\alpha, \beta \in \mathbf{N}_{+}$, and $\alpha+\beta=x$.
Subtracting the two equations gives $5 \times 2^{\alpha}=2\left(2^{\beta-1}+1\right)$, i.e., $5 \times 2^{\alpha-1}=2^{\beta-1}+1$.
Since $5 \mid\left(2^{\beta-1}+1\right)$, we have $\beta \geqslant 3$.
Thus, $2 \times\left(2^{\beta-1}+1\right)$.
Hence, $\alpha=1, \beta=3$.
Therefore, $x=\alpha+\beta=4$.
(2) If $1+5 \times 2^{x}$ is the square of a fraction, then $x \in \mathbf{Z}_{\text {. }}$.
Let $x=-y\left(y \in \mathbf{N}_{+}\right)$. Then
$1+5 \times 2^{x}=\frac{2^{y}+5}{2^{y}}$.
Since $2 \times\left(2^{y}+5\right)$, we have $2 \mid y$.
Let $y=2 y_{1}, 2^{y}+5=m^{2}\left(m \in \mathbf{N}_{+}\right)$. Then $\left(m+2^{y_{1}}\right)\left(m-2^{y_{1}}\right)=5$.
Therefore, $\left\{\begin{array}{l}m+2^{y_{1}}=5, \\ m-2^{y_{1}}=1 .\end{array}\right.$
Subtracting the two equations gives $2^{y_{1}+1}=4$. Hence, $y_{1}=1$.
Therefore, $y=2 y_{1}=2, x=-y=-2$.
In summary, $x=-2$ or 4.