Let's find a perfect square consisting of 4 digits, knowing that the number formed by the first two digits exceeds the number formed by the last two digits by one.
Let $x$ denote the two-digit number formed by the last two digits, the equation solving the problem is:

$$
100(x+1)+x=y^{2}
$$

which can also be written as:

$$
101 x=y^{2}-100=(y+10)(y-10)
$$

Since 101 is a prime number, either $y+10$ or $y-10$ must be divisible by 101. However, since $y^2$ is a four-digit number, $y$ can have at most two digits. The only possible assumption, therefore, is that

$$
y+10=101
$$

and from this

$$
y=91
$$

whose square is: 8281, which satisfies the problem.