Example 2 In the sequence $\left\{a_{n}\right\}$,
$$
a_{1}=2, a_{n+1}=4 a_{n}-3 n+1\left(n \in \mathbf{N}_{+}\right) \text {. }
$$

Find the general term formula of the sequence $\left\{a_{n}\right\}$.
Given $a_{n+1}=4 a_{n}-3 n+1$, let
$$
\begin{array}{l}
a_{n+1}-[a(n+1)+b] \\
=4\left[a_{n}-(a n+b)\right] .
\end{array}
$$

Then $a(n+1)+b-4(a n+b)=-3 n+1$.
We get $a=1, b=0$.
Thus, $a_{n+1}-(n+1)=4\left(a_{n}-n\right)$.
Also, $a_{1}-1=1 \neq 0$, so the sequence $\left\{a_{n}-n\right\}$ is a geometric sequence with the first term $a_{1}-1=1$ and common ratio 4.

Therefore, $a_{n}-n=4^{n-1}$, which means the general term formula of the sequence $\left\{a_{n}\right\}$ is
$$
a_{n}=4^{n-1}+n \text {. }
$$