Let $x_{0}+\sqrt{2003} y_{0}$ be the fundamental solution of the Pell equation
$$
x^{2}-2003 y^{2}-1
$$

Find the solution $(x, y)$ of (1), where $x, y$ are positive numbers and all prime factors of $x$ divide $x_{0}$.
Since $x_{0}+\sqrt{2003} y_{0}$ is the fundamental solution, all solutions $x+\sqrt{2003 y} y=\left(x_{0}+\right.$ $\left.\sqrt{2003} y_{0}\right)^{n}$.
When $n=2 k$ ( $k$ is a positive integer),
$$
x=\sum_{m=1}^{k} x_{0}^{2 m} \cdot 2003^{k-m} y_{0}^{2 k-2 m} \mathrm{C}_{2 k}^{2 m}+2003^{k} y_{0}^{2 k},
$$

Since $\left(x_{0}, y_{0}\right)$ satisfies (1), $x_{0}$ is coprime with 2003 and $y_{0}$. From (2), if a prime factor of $x$ divides $x_{0}$, then this prime factor must divide $2003 y_{0}$. This is impossible.
When $n=2 k+1$ ( $k$ is a positive integer),
$$
x=\sum_{m=1}^{k} x_{0}^{2 m+1} \cdot 2003^{k-m} \cdot y_{0}^{2 k-2 m} C_{2 k+1}^{2 m}+(2 k+1) x_{0} \cdot 2003^{k} y_{0}^{2 k} .
$$

Let $p$ be a prime factor of $x$, and also a prime factor of $x_{0}$. The number of times $p$ appears in $2 m$ is clearly less than $2 m$, i.e., the number of times in $x_{0}^{2 m}$. Therefore, the number of times $p$ appears in $x_{0}^{2 m+1} \mathrm{C}_{2 k+1}^{2 m}=x_{0}(2 k+1) \cdot x_{0}^{2 m} \cdot \frac{1}{2 m} \mathrm{C}_{2 k}^{2 m-1}$ is greater than the number of times $p$ appears in $x_{0}(2 k+1)$.

Thus, the number of times $p$ appears on the right side of (4) is the same as the number of times $p$ appears in $(2 k+1) x_{0}$. Since the number of times each prime factor of $x$ appears is equal to the number of times this prime factor appears in $(2 k+1) x_{0}$, it follows that $x \leqslant(2 k \mid 1) x_{0}$. This clearly contradicts (3). Therefore, the solution $(x, y)$ that meets the requirements can only be the fundamental solution itself, i.e., $x=x_{0}, y=y_{0}$.