The ordered pair of four-digit numbers $(2025, 3136)$ has the property that each number in the pair is a perfect square and each digit of the second number is $1$ more than the corresponding digit of the first number. Find, with proof, all ordered pairs of five-digit numbers and ordered pairs of six-digit numbers with the same property: each number in the pair is a perfect square and each digit of the second number is $1$ more than the corresponding digit of the first number.
To solve this problem, we need to find ordered pairs of five-digit and six-digit numbers \((a^2, b^2)\) such that each number in the pair is a perfect square and each digit of \(b^2\) is 1 more than the corresponding digit of \(a^2\).

### Five-Digit Numbers

1. **Set up the equation:**
   \[
   a^2 - b^2 = 11111
   \]
   This can be factored using the difference of squares:
   \[
   (a - b)(a + b) = 11111
   \]

2. **Factorize 11111:**
   \[
   11111 = 41 \times 271
   \]
   Therefore, we have two possible factorizations:
   \[
   a - b = 41 \quad \text{and} \quad a + b = 271
   \]

3. **Solve the system of equations:**
   \[
   a - b = 41
   \]
   \[
   a + b = 271
   \]
   Adding these two equations:
   \[
   2a = 312 \implies a = 156
   \]
   Subtracting the first equation from the second:
   \[
   2b = 230 \implies b = 115
   \]

4. **Verify the solution:**
   \[
   a^2 = 156^2 = 24336
   \]
   \[
   b^2 = 115^2 = 13225
   \]
   Check if each digit of \(b^2\) is 1 more than the corresponding digit of \(a^2\):
   \[
   24336 \quad \text{and} \quad 13225
   \]
   Indeed, each digit of 13225 is 1 more than the corresponding digit of 24336.

Thus, the ordered pair of five-digit numbers is:
\[
\boxed{(24336, 13225)}
\]

### Six-Digit Numbers

1. **Set up the equation:**
   \[
   a^2 - b^2 = 111111
   \]
   This can be factored using the difference of squares:
   \[
   (a - b)(a + b) = 111111
   \]

2. **Factorize 111111:**
   \[
   111111 = 3 \times 7 \times 11 \times 13 \times 37
   \]
   We need to find pairs \((a - b)\) and \((a + b)\) such that their product is 111111 and they satisfy the conditions for \(a\) and \(b\) being integers.

3. **Check possible pairs:**
   We need to find pairs \((a - b)\) and \((a + b)\) such that:
   \[
   a^2 < 1000000 \implies a \leq 999
   \]
   \[
   b^2 > 99999 \implies b \geq 317
   \]
   This implies:
   \[
   635 \leq a + b \leq 1997
   \]

4. **Test possible pairs:**
   We test the pairs:
   \[
   (a - b, a + b) = (3, 37037), (7, 15873), (11, 10101), (13, 8547), (21, 5291), (33, 3367), (37, 3003), (39, 2847), (77, 1443), (91, 1221), (111, 1001), (143, 777)
   \]
   For each pair, solve for \(a\) and \(b\):
   \[
   a = \frac{(a + b) + (a - b)}{2}
   \]
   \[
   b = \frac{(a + b) - (a - b)}{2}
   \]

5. **Verify each solution:**
   After testing each pair, none of the pairs satisfy the condition that each digit of \(b^2\) is 1 more than the corresponding digit of \(a^2\).

Thus, there are no six-digit numbers that satisfy the given conditions.