Example 6 Find the smallest positive integer $d_{0}$ such that $7^{d} \equiv 1\left(\bmod 2^{r}\right)$ (integer $\left.r \geqslant 3\right)$.
Solution: $\because \Phi\left(2^{r}\right)=2^{r}\left(1-\frac{1}{2}\right)=2^{r-1}$ and $d_{0} \mid \Phi(2r)$, $\therefore d_{0}=2^{k}, 0 \leqslant k \leqslant r-1$.

First, we prove that for any odd number $a$, there must be $a^{2^{-7}} \equiv 1\left(\bmod 2^{r}\right)$: By induction, let $a=2 t+1$,

When $r=3$, $a^{2}=4 t(t+1)+1 \equiv 1\left(\bmod 2^{3}\right)$, assume when $r=n$, $a^{2^{2-2}} \equiv 1\left(\bmod 2^{n}\right)$, then when $r=n+1$, $a^{2^{-1}}-1=\left(a^{2^{-x}}-1\right)\left(a^{2^{n-2}}+1\right)$ is divisible by $2^{n+1}$, i.e., $a^{2^{2-1}} \equiv 1\left(\bmod 2^{n+1}\right)$, so the conclusion holds.

From this, we know that $k$ in $d_{0}=2^{k}$ satisfies $0 \leqslant k \leqslant r-2$. We prove: For any integer $r \geqslant 3$, there must be $7^{2^{-3}} \neq 1 \pmod{2r}$: By induction, when $r=3$, it is obviously true. Assume when $r=n$, $7^{2^{n-3}} \not \equiv 1\left(\bmod 2^{n}\right)$, $\because 7^{2^{n-3}} \equiv 1\left(\bmod 2^{n-1}\right)$,
$\therefore 7^{2^{n-3}}=1+s \cdot 2^{n-1}$, where the integer $2 Y_{s}$,
$\therefore 7^{2^{n-2}}=\left(7^{2^{n-3}}\right)^{2}=1+s\left(1+s \cdot 2^{n-2}\right) 2^{n}$, $2 \times 1+s \cdot 2^{n-2}$,

i.e., $7^{2^{n-2}} \neq 1\left(\bmod 2^{n+1}\right)$, $\therefore 7^{2^{r^{-3}}} \neq 1 \pmod{2r}$, from this we deduce: $7^{2} \neq 1\left(\bmod 2^{r}\right), 0 \leqslant j \leqslant r-3$, hence $d_{0}=2^{-2}$ is the desired result.