Example 5 Find all positive integers $x, y$ such that
$$y^{x}=x^{50}$$
Let $x, y$ be positive integers satisfying the conditions, and let $x=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$ be the prime factorization of $x$. Then, since $y$ is a positive integer, for $1 \leqslant i \leqslant k$, we have $x \mid 50 \alpha_{i}$. Now, let's first discuss the prime factors of $x$.

If $x$ has a prime factor $p$ different from 2 and 5, and let $p^{\alpha} \| x$, then from the previous result, we know $x \mid 50 \alpha$, which implies $p^{\alpha} \mid 50 \alpha$. Since $p \neq 2, 5$, it follows that $p^{\alpha} \mid \alpha$. However, for any prime $p$ and positive integer $\alpha$, we have $p^{\alpha} > \alpha$, so $p^{\alpha} \mid \alpha$ cannot hold. This indicates that the prime factors of $x$ can only be 2 or 5.

Thus, we can set $x=2^{\alpha} \cdot 5^{\beta}$ (where $\alpha, \beta$ are non-negative integers). In this case, $x \mid 50 \alpha, x \mid 50 \beta$, so $2^{\alpha} \mid 50 \alpha, 5^{\beta} \mid 50 \beta$. The former requires $2^{\alpha-1} \mid \alpha$, and the latter requires $5^{\beta-2} \mid \beta$. Note that when $\alpha \geqslant 3$, $2^{\alpha-1} > \alpha$, and when $\beta \geqslant 3$, $5^{\beta-2} > \beta$. Therefore, $0 \leqslant \alpha \leqslant 2, 0 \leqslant \beta \leqslant 2$. This indicates that $x$ can only take the values $1, 2, 2^{2}, 5, 5^{2}, 2 \times 5, 2^{2} \times 5, 2 \times 5^{2}, 2^{2} \times 5^{2}$.

Substituting the above values of $x$ into equation (1), we can obtain all solutions as $(x, y) = (1,1)$, $\left(2,2^{25}\right), \left(2^{2}, 2^{25}\right), \left(5,5^{10}\right), \left(5^{2}, 5^{4}\right), \left(10,10^{5}\right), (50,50), (100,10)$, a total of 8 solutions.