Consecutive Integers whose Sums of Squares of Divisors are Equal

Theorem
The only two consecutive positive integers whose sums of the squares of their divisors are $6$ and $7$.

Proof
The divisors of $6$ are
 * $1, 2, 3, 6$

and so the sum of the squares of the divisors of $6$ is:
 * $1^2 + 2^2 + 3^2 + 6^2 = 1 + 4 + 9 + 36 = 50$

The divisors of $7$ are
 * $1, 7$

and so the sum of the squares of the divisors of $7$ is:
 * $1^2 + 7^2 = 1 + 49 = 50$

It remains to be shown that there are no more: