II. (25 points) Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.

Arrange all positive integers that satisfy the following conditions in descending order, denoted as $M$, and the $k$-th number as $b_{k}$: each number's any three consecutive digits form a non-zero perfect square. If $b_{16}-b_{20}=2^{n}$, find $n$.
II. Notice that, among three-digit numbers, the perfect squares are $100, 121, 144, 169, 196, 225, 256, 289$, $324, 361, 400, 441, 484, 529, 576, 625$, $676, 729, 784, 841, 900, 961$.
For each number in $M$, consider its leftmost three digits.
(1) The six-digit number with the leftmost three digits 100 is 100169, the five-digit numbers are $10016, 10049$, and the four-digit numbers are $1001, 1004, 1009$.
(2) The six-digit number with the leftmost three digits 400 is 400169, the five-digit numbers are 40016, 40049, and the four-digit numbers are $4001, 4004, 4009$.
(3) The six-digit number with the leftmost three digits 900 is 900169, the five-digit numbers are $90016, 90049$, and the four-digit numbers are $9001, 9004, 9009$.
(4) When the leftmost three digits are $144, 196, 225, 484, 625, 784$, the corresponding four-digit numbers are $1441, 1961, 2256, 4841, 6256, 7841$.
(5) The rest are all three-digit numbers. After sorting the above numbers in descending order, we find that $b_{16}=4009, b_{20}=1961$.
Thus, $b_{16}-b_{20}=2048=2^{11}$, which means $n=11$.