Example 2 Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n}=2^{2 n} a_{n-1}+2^{n^{2}} n \quad (n=2,3, \cdots) \text {. }
$$

Find the general term $a_{n} \ (n=1,2, \cdots)$.
(Sixth Northern Mathematical Olympiad Invitational Competition)
Let $b_{n}=\frac{a_{n}}{2^{n^{2}}}$. Then $b_{n}=2 b_{n-1}+n$, i.e., $b_{n}+n=2\left(b_{n-1}+n-1\right)+2, b_{1}=1$.
Let $c_{n}=b_{n}+n$. Then $c_{n}=2 c_{n-1}+2$, i.e.,
$$
\begin{array}{l}
c_{n}+2=2\left(c_{n-1}+2\right), c_{1}=2 \\
\Rightarrow c_{n}+2=2^{n-1}\left(c_{1}+2\right)=2^{n+1} \\
\Rightarrow c_{n}=2^{n+1}-2 .
\end{array}
$$

Therefore, $b_{n}=c_{n}-n=2^{n+1}-n-2$.
Thus, $a_{n}=2^{n^{2}} b_{n}$
$$
=2^{n^{2}}\left(2^{n+1}-n-2\right)(n=1,2, \cdots) .
$$