Since 257 is a prime number, we have
$$\begin{aligned} \left(\frac{34}{257}\right) & =\left(\frac{2}{257}\right)\left(\frac{17}{257}\right)=\left(\frac{17}{257}\right) \\ & =\left(\frac{257}{17}\right)=\left(\frac{2}{17}\right)=1 \end{aligned}$$

Therefore, the original congruence equation must have a solution.
Since $257=4 \times 2^{6}+1$, we have $n=2^{6}, \lambda=6, \mu=1$, $\frac{u+1}{2}=1$, so $a^{u}=34$. We also have
$$\begin{array}{c} 34^{2}=1156 \equiv 128(\bmod 257) \\ 34^{2^{2}} \equiv 128^{2}=16384 \equiv 193 \equiv-64(\bmod 257) \\ 34^{2^{3}} \equiv(-64)^{2}=4096 \equiv 241 \equiv-16(\bmod 257) \\ 34^{2^{4}} \equiv(-16)^{2}=256 \equiv-1 \quad(\bmod 257) \end{array}$$

Thus,
$$\mu=4, \lambda-\mu=6-4=2,$$

Hence,
$$h=2^{2} \cdot t=4 t, \quad t \text { is odd }$$

And we have $\square$
$$1 \leqslant t \leqslant 2^{4}-1=15$$

Since $257=12 \times 21+5$, 3 is a quadratic non-residue modulo 257, so we have
$$\left(b^{2 u}\right)^{h}=\left(3^{2}\right)^{4 t}=\left(3^{8}\right)^{t}$$

However, we have
$$\begin{array}{c} 3^{8}=9^{4}=81^{2}=6561 \equiv 136 \equiv-121(\bmod 257) \\ \left(3^{8}\right)^{3} \equiv(-121)^{3}=(14641)(-121) \equiv(-8)(-121) \\ \equiv 968 \equiv 197 \equiv-60(\bmod 257) \\ \left(3^{8}\right)^{5} \equiv(-8)(-60)=480 \equiv-34(\bmod 257) \end{array}$$

Therefore, we must have $t=5$, so $h=4 t=4 \times 5=20$,
$$n-h=2^{6}-20=44,$$

Since
$$\left(3^{8}\right)^{5} \equiv-34(\bmod 257)$$

We have
$$\begin{aligned} 3^{44} & \equiv 3^{4} \times 3^{40} \equiv 81 \times(-34) \equiv-2754 \\ & \equiv-184 \equiv 73(\bmod 257) \end{aligned}$$

And
$$34 \times 73=2482 \equiv 169 \equiv-88(\bmod 257)$$

Thus, by (10), the solution to the original equation is
$$x \equiv \pm 88(\bmod 257)$$

For the case of $p^{\alpha}$ modulo, where $p>2, \alpha>1$.
We already know how to determine whether
$$x^{2} \equiv a(\bmod p),(a, p)=1$$

has a solution, and if it does, how to find it. In this section, we will discuss, if $p$ is an odd prime, $\alpha$ is an integer greater than 1, the congruence equation
$$x^{2} \equiv a\left(\bmod p^{\alpha}\right),(\alpha, p)=1$$