Let's find all the square numbers that, when written in binary, consist entirely of 1s.
If a binary number ends with 2 ones, then the remainder when divided by 4 is 3. The other place values, $2^{2}, 2^{3}, 2^{4}$, etc., are indeed divisible by 4. It is easy to see that there is no square number that leaves a remainder of 3 when divided by 4. Therefore, a square number cannot end with 2 ones in binary form. But then we are left with only one possibility, that the binary number is 1, and this is indeed a square number.