Example 3 Let $a$ be a prime number, $b$ be a positive integer, and $9(2 a+b)^{2}=509(4 a+511 b)$. Find the values of $a$ and $b$.

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Solution: Since $9(2 a+b)^{2}=3^{2}(2 a+b)^{2}$ is a perfect square, therefore, $509(4 a+511 b)$ is a perfect square. Since 509 is a prime number, we can let
$$
4 a+511 b=509 \times 3^{2} k^{2} \text {. }
$$

Thus, the original equation becomes
$$
9(2 a+b)^{2}=509^{2} \times 3^{2} k^{2} \text {, }
$$

which means $2 a+b=509 k$.
Therefore, $b=509 k-2 a$. Substituting into equation (1) gives
$$
4 a+511(509 k-2 a)=509 \times 3^{2} k^{2} \text {. }
$$

Solving for $a$ yields $a=\frac{k(511-9 k)}{2}$.
Since $a$ is a prime number, i.e., $\frac{k(511-9 k)}{2}$ is a prime number, there are the following cases:
(1) When $k=1$, $a=\frac{k(511-9 k)}{2}=\frac{511-9}{2}$ $=251$ is a prime number, which meets the condition. At this time,
$$
b=509 k-2 a=509-502=7 \text {. }
$$
(2) When $k=2$, $a=\frac{k(511-9 k)}{2}=511-$ $18=493=17 \times 29$ is not a prime number, so it is discarded.
(3) When $k>2$ and $k$ is odd, since $a=$ $\frac{k(511-9 k)}{2}=k \cdot \frac{511-9 k}{2}$ is a prime number, and $k>1$, so $\frac{511-9 k}{2}=1$. But $\frac{511-9 k}{2}=1$ has no integer solution, so it is discarded.
(4) When $k>2$ and $k$ is even, since $a=$ $\frac{k(511-9 k)}{2}=\frac{k}{2}(511-9 k)$ is a prime number, and $\frac{k}{2}>$ 1, so $511-9 k=1$. But $511-9 k=1$ has no integer solution, so it is discarded.
In summary, $a=251, b=7$.