$2 \cdot 82$ Find the largest integer $x$ such that $4^{27}+4^{1000}+4^{x}$ is a perfect square.
［Solution］Since
$$
4^{27}+4^{1000}+4^{x}=2^{54}\left(1+2 \cdot 2^{1945}+2^{2 x-54}\right)
$$

Therefore, when $2 x-54=2 \cdot 1945$, i.e., when $x=1972$, $4^{27}+4^{1000}+4^{x}$ is a perfect square.
On the other hand, when $x>1972$, since
$$
2^{2(x-27)}<1+2 \cdot 2^{1945}+2^{2 \cdot(x-27)}<\left(2^{x-27}+1\right)^{2},
$$

$1+2 \cdot 2^{1945}+2^{2 x-54}$ lies between the squares of two consecutive natural numbers, hence it is not a perfect square, and the given expression $4^{27}+4^{1000}+4^{x}$ is also not a perfect square.
Thus, the maximum integer $x$ is 1972.