Solution: Let $y_{n}=\sqrt{b^{2}+4 a x_{n}}$, then
$$y_{0}=b, x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a} \text {. }$$

Thus, we have $y_{n+1}^{2}=\left(y_{n}+2 a\right)^{2}$.
Since $y_{n}>0$, then $y_{n+1}-y_{n}=2 a$, hence $\left\{y_{n}\right\}$ is an arithmetic sequence with a common difference of $2 a$.
From this, we deduce $y_{n+1}=y_{0}+2 a n=2 a n+b$, so
$$x_{n}=\frac{y_{n}^{2}-b^{2}}{4 a}=a n^{2}+b n \text {. }$$

Comment: This problem transforms the original irrational recurrence relation into a familiar arithmetic sequence by expressing $x_{n}$ in terms of $y_{n}$ through a substitution. If the terms inside and outside the square root are both linear in $x_{n}$, when the coefficients change, a similar substitution can transform it into the form $y_{n+1}=p y_{n}+q$, first solving for $y_{n}$, then for $x_{n}$.