3. Let $n$ and $m$ be positive integers of different parity, and $n > m$. Find all integers $x$ such that $\frac{x^{2^{n}}-1}{x^{2^{m}}-1}$ is a perfect square.
(Pan Chengdu)
3. Let $\frac{x^{2^{n}}-1}{x^{2^{m}}-1}=A^{2}$, where $A \in \mathbf{N}$, then
$$
A^{2}=\prod_{i=m}^{n-1}\left(x^{2^{4}}+1\right) \text {. }
$$

Notice that for $i \neq j,\left(x^{2^{2}}+1, x^{2^{2}}+1\right)=\left\{\begin{array}{ll}1, & 2 \mid x, \\ 2, & 2 \nmid x .\end{array}\right.$
Case 1: If $2 \mid x$, then $\forall m \leqslant i \leqslant n-1$, the number $x^{2^{i}}+1$ is a perfect square, so it can only be $x=0$;

Case 2: If $2 \nmid x$, then $2^{n-m} \prod_{i=m}^{n-1}\left(\frac{x^{2^{1}}+1}{2}\right)$. And $\left(\frac{x^{2^{4}}+1}{2}, \frac{x^{2^{j}}+1}{2}\right)=1$, so it must be that $2 \mid (n-m)$, which is a contradiction. In summary, the only integer $x$ that satisfies the condition is $x=0$.