Let this number be $n = \overline{aabb} = \overline{a0b} \times 11$.

If $n$ is a square,  then it must have even number of divisor $11$. Hence, it must hold that $11 \mid \overline {a0b}$. By the division by $11$ property, we must have $11 \mid a + b$. Therefore, there are $8$ possibilities:$$(a, b) = (2, 9), (9, 2), (3, 8), (8, 3), (4, 7), (7, 4), (5, 6), (6, 5)$$

By square number's ending digit property, only $(a, b)=$ $(2, 9)$, $(7,4)$, $(5, 6)$, and $(6, 5)$ are possible. By # 4149, the last two digits of a square cannot be both odd. Therefore, neither $(2, 9)$ nor $(6, 5)$ is possible. Additionally, by # 4150, $(5,6)$ cannot be a square either. This leaves $(7, 4)$ the only possible choice.

Because $\boxed{7744}=88^2$, there it is the only solution to this problem.