6th and 7th divisor of integer

Question

Is there a positive integer $n$ such that $n=d_6 ^ 2 + d_7 ^ 2 -1$, where $d_k$ is $k^{th}$ divisor of $n$? ( $1$ is first, etc., $n$ is last) So far have only concluded that $n$ in that case should have more than $12$ divisors, since $n=d_6 ^ 2 + d_7 ^ 2 -1 > d_6 d_7$ and $d_6$and $d_7$ are not in the middle of array of divisors of $n$. Also, if $d_6$ and $d_7$ are relatively prime, then we would need to solve equation $d_6 d_7 m = d_6 ^ 2 + d_7 ^ 2 -1$, but they are not necessarily relatively prime. Thanks for reading, any help is appreciated. (problem came from one friend)