Example 3 Try to find a positive integer $k$ other than 1, such that $k$ and $k^{4}$ can both be expressed as the sum of squares of two consecutive integers, and prove that such a $k$ is unique.

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$$
\begin{array}{l}
\text { Solution: Let } k=n^{2}+(n+1)^{2} . \\
\text { By }\left(a^{2}+b^{2}\right)^{2}=\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2}, \text { we have } \\
k^{2}=(2 n+1)^{2}+[2 n(n+1)]^{2}, \\
k^{4}=\left(4 n^{4}+8 n^{3}-4 n-1\right)^{2}+ \\
\quad\left(8 n^{3}+12 n^{2}+4 n\right)^{2} . \\
\text { Let }\left|\left(4 n^{4}+8 n^{3}-4 n-1\right)-\left(8 n^{3}+12 n^{2}+4 n\right)\right| \\
=1 \text {, i.e., }| 4 n^{4}-12 n^{2}-8 n-1 \mid=1 . \\
\text { If } 4 n^{4}-12 n^{2}-8 n-1=1 \text {, then } \\
4 n^{4}-12 n^{2}-8 n=2,
\end{array}
$$

The left side is divisible by 4, but the right side is not, which is a contradiction;
If $4 n^{4}-12 n^{2}-8 n-1=-1$, then
$$
4 n(n+1)^{2}(n-2)=0 \text {, }
$$

Thus, $n=-1,0,2$.
When $n=-1$ or 0, $k=1$, which contradicts the given condition;
When $n=2$, $k=13$, in this case we have
$$
13=2^{2}+3^{2}, 13^{4}=119^{2}+120^{2} \text {, }
$$
$k$ is unique.
This method uses the formula from $k$ to $k^{4}$,
$$
\left(a^{2}+b^{2}\right)^{2}=\left(a^{2}-b^{2}\right)^{2}+(2 a b)^{2} \text {, }
$$

We can also use the solution formula for Pythagorean equations to prove from $k^{4}$ to $k$, but the process is more complicated and will not be detailed here.
$$