$2 \cdot 87$ Let $4^{27}+4^{500}+4^{n}$ be a perfect square (square of an integer), find the maximum value of the integer $n$.


[Solution] Factorization yields
$$\begin{aligned}
& 4^{27}+4^{500}+4^{n} \\
= & 4^{27}\left(4^{n-27}+4^{473}+1\right) \\
& 4^{n-27}+4^{473}+1=\left(2^{n-27}\right)^{2}+2 \cdot 2^{945}+1
\end{aligned}$$

When $n-27=945$, i.e., $n=972$, (2) is a perfect square, and thus (1) is a perfect square.

When $n>972$,
$$\begin{aligned}
& \left(2^{n-27}+1\right)^{2} \\
= & \left(2^{n-27}\right)^{2}+2 \cdot 2^{n-27}+1 \\
> & \left(2^{n-27}\right)^{2}+2 \cdot 2^{945}+1 \\
> & \left(2^{n-27}\right)^{2}
\end{aligned}$$

Therefore, at this time (2) is not a perfect square, and thus (1) is not a perfect square. Hence, the maximum value of $n$ is 972.