Example 8 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 $b$.
(2008, National Junior High School Mathematics Competition)
The original equation can be transformed into
$$
\left(\frac{6 a+3 b}{509}\right)^{2}=\frac{4 a+511 b}{509} \text {. }
$$

Let $m=\frac{6 a+3 b}{509}, n=\frac{4 a+511 b}{509}$.
Then $3 n-511 m+6 a=0$.
Also, $n=m^{2}$, so
$$
3 m^{2}-511 m+6 a=0 \text {. }
$$

From the original equation, we know that $(2 a+b)^{2}$ is divisible by the prime number 509, hence $2 a+b$ is divisible by 509. Therefore, $m$ is an integer, meaning the quadratic equation in $m$
$$
3 m^{2}-511 m+6 a=0
$$

has integer roots.
Thus, its discriminant $\Delta=511^{2}-72 a$ must be a perfect square.
Assume $\Delta=511^{2}-72 a=k^{2}$ (where $k$ is a natural number). Then $72 a=511^{2}-k^{2}=(511+k)(511-k)$.
Since $511+k$ and $511-k$ have the same parity, and $511+k \geqslant 511$, there are only a few possible cases.
Adding the first four systems of equations respectively gives
$$
\begin{array}{l}
36 a+2=1022,18 a+4=1022, \\
12 a+6=1022,6 a+12=1022,
\end{array}
$$

none of which have integer solutions.
Adding the last two systems of equations respectively gives
$4 a+18=1022$, solving for $a=251$;
$2 a+36=1022$, solving for $a=493$.
However, $493=17 \times 29$ is not a prime number, so it is discarded.
In summary, $a=251$. At this point, the solution to equation (1) is $m=3$ or $m=\frac{502}{3}$ (discarded).

Furthermore, $b=\frac{509 \times 3-6 \times 251}{3}=7$.