Example 9 Find the largest positive integer $n$, such that $4^{27}+4^{50}+4^{n}$ is a perfect square.



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Example 9 Find the largest positive integer $n$, such that $4^{27}+4^{50}+4^{n}$ is a perfect square.
Solution: When $n>27$,
$$
A=4^{27}+4^{500}+4^{n}=4^{27}\left(1+4^{473}+4^{n-27}\right) \text {. }
$$

Since $4^{27}$ is a perfect square, for $A$ to be a perfect square, $1+4^{473}+4^{n-27}$ must be a perfect square.
Notice that
$$
\begin{array}{l}
1+4^{473}+4^{n-27}=1+2^{2 \times 473}+\left(2^{n-27}\right)^{2} \\
=1+2 \times 2^{2 \times 473-1}+\left(2^{n-27}\right)^{2},
\end{array}
$$

Thus, when $n-27=2 \times 473-1$, i.e., $n=972$,
$$
\begin{array}{l}
1+4^{473}+4^{n-27} \\
=1+2 \times 2^{2 \times 473-1}+\left(2^{2 \times 473-1}\right)^{2} \\
=\left(1+2^{2 \times 473-1}\right)^{2}
\end{array}
$$

is a perfect square. At this point, $A$ is a perfect square.
Therefore, $n=972$ meets the condition.
If $n>972$, then $2 \times 473-1\left(2^{n-27}\right)^{2},
\end{array}
$$

then
$$
\left(2^{n-27}\right)^{2}<1+4^{4 / 3}+4^{n-27}<\left(1+2^{n-27}\right)^{2} \text {. }
$$

Hence, $1+4^{473}+4^{n-27}$ is not a perfect square, which is a contradiction.
Therefore, $n \leqslant 972$.
In conclusion, $n_{\max }=972$.