6. (2004 Croatian Mathematical Competition) Find all positive integers with more than two digits such that each pair of adjacent digits forms a square integer.

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The translation maintains the original format and line breaks as requested. However, it's worth noting that "square integer" might be better phrased as "perfect square" in this context, which is more commonly used in English mathematical literature. If you prefer, I can adjust the translation accordingly.
6. It is known that the squares of positive integers with two digits are: $16,25,36,49,64,81$.

Notice that, starting from the given numbers, there is at most one two-digit square, so after the first two-digit number is chosen, the remaining part of the desired number is uniquely determined. Since there are no two-digit square numbers starting with 5 or 9, the desired number cannot start with 25 or 49.
From 16, we get 164,1649;
From 36, we get 364,3649;
From 64, we get 649;
From 81, we get $816,8164,81649$.
Therefore, the numbers that satisfy the condition are
$$
164,1649,364,3649,649,816,8164,81649 .
$$