The simplest method is to consider the dual form of $(\sqrt{2}+1)^{n}$, which is $(\sqrt{2}-1)^{n}$.
By the binomial theorem, we have
$$\begin{array}{l} (\sqrt{2}+1)^{n}=\sqrt{a}+\sqrt{b} \\ \Rightarrow(\sqrt{2}-1)^{n}=\sqrt{a}-\sqrt{b}\left(a, b \in \mathbf{N}_{+}\right) \\ \Rightarrow(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b}) \\ =(\sqrt{2}-1)^{n}(\sqrt{2}+1)^{n}=1 \\ \Rightarrow a=m, b=m-1 . \end{array}$$

Proof:
(1) When $n=1, m=2$, it is clearly true.
(2) Assume that when $n=k$,
$$(\sqrt{2}+1)^{k}=\sqrt{m}+\sqrt{m-1} \text {. }$$

Then when $n=k+1$,
$$\begin{array}{l} (\sqrt{2}+1)^{k+1}=(\sqrt{2}+1)^{k}(\sqrt{2}+1) \\ =(\sqrt{m}+\sqrt{m-1})(\sqrt{2}+1) \\ =(\sqrt{m}+\sqrt{2 m-2})+(\sqrt{2 m}+\sqrt{m-1}) \\ =\sqrt{3 m-2+2 \sqrt{2 m(m-1)}}+ \\ \quad \sqrt{3 m-1+2 \sqrt{2 m(m-1)}} . \end{array}$$

Next, we need to prove that $\sqrt{2 m(m-1)} \in \mathbf{N}_{+}$.
By the binomial theorem, we have
$$(\sqrt{2}+1)^{k}=a+b \sqrt{2}\left(a, b \in \mathbf{N}_{+}\right) \text {. }$$

Also,
$$a+b \sqrt{2}=\sqrt{m}+\sqrt{m-1},$$

so, one of $m$ and $m-1$ is a perfect square, and the other is twice a perfect square.
Therefore, $2 m(m-1)$ is a perfect square, i.e., $\sqrt{2 m(m-1)} \in \mathbf{N}_{+}$.
Thus, when $n=k+1$, the conclusion holds.
Combining (1) and (2), the conclusion is established.
From the above solution process, we see that $m$ is a sequence of numbers determined by $n$. If we set each $m$ corresponding value to $a_{n}$, then what is the general term formula of $\left\{a_{n}\right\}$?

It is not difficult to find that when $n$ changes from $k$ to $k+1$, $m$ changes to $3 m-1+2 \sqrt{2 m(m-1)}$.
Thus, we have the nonlinear recurrence relation
$$a_{n+1}=3 a_{n}-1+2 \sqrt{2 a_{n}\left(a_{n}-1\right)}\left(a_{1}=2\right) .$$

This recurrence relation can be transformed into a linear relation through the following operations.
$$\begin{array}{l} \text { By } a_{n+1}=3 a_{n}-1+2 \sqrt{2 a_{n}\left(a_{n}-1\right)} \\ \Rightarrow\left(a_{n+1}-3 a_{n}+1\right)^{2}=4 \times 2 a_{n}\left(a_{n}-1\right) \\ \Rightarrow a_{n+1}^{2}+a_{n}^{2}+2 a_{n+1}+2 a_{n}-6 a_{n+1} a_{n}+1=0 \\ \Rightarrow a_{n-1}^{2}+a_{n}^{2}+2 a_{n-1}+2 a_{n}-6 a_{n} a_{n-1}+1 \\ \quad=0(n \geqslant 2) \text {. } \end{array}$$

Comparing equations (1) and (2), we know that $a_{n-1}$ and $a_{n+1}$ are the two roots of the equation
$$x^{2}+\left(2-6 a_{n}\right) x+a_{n}^{2}+2 a_{n}+1=0$$

Thus, $a_{n+1}+a_{n-1}=6 a_{n}-2$.
Rearranging, we get
$a_{n+1}-\frac{1}{2}=6\left(a_{n}-\frac{1}{2}\right)-\left(a_{n-1}-\frac{1}{2}\right)$.
Let $b_{n}=a_{n}-\frac{1}{2}$.
Then $b_{n+1}=6 b_{n}-b_{n-1}$, where $b_{1}=\frac{3}{2}, b_{2}=\frac{17}{2}$.
The characteristic equation is $x^{2}=6 x-1$.
Solving, we get $x=3 \pm 2 \sqrt{2}$.
Thus, $b_{n}=A(3+2 \sqrt{2})^{n}+B(3-2 \sqrt{2})^{n}$.
Substituting $b_{1}$ and $b_{2}$, we get $A=B=\frac{1}{4}$.
Therefore, $a_{n}=\frac{1}{4}(3+2 \sqrt{2})^{n}+\frac{1}{4}(3-2 \sqrt{2})^{n}+\frac{1}{2}$.