It's not as easy as similar ones!

From the equality $n = d_6^2 +d_7^2 -1$ we see that $d_6$ and $d_7$ are relatively prime and $d_7 | d_6^2 -1 = (d_6-1)(d_6+1)$ , $d_6 | d_7^2 -1 = (d_7-1)(d_7+1)$ . Suppose that $d_6 = ab$ , $d_7 = cd$ with $1 < a < b$ , $1 < c < d$ . Then $n$ has $7$ divisors smaller than $d_7$ , namely $1,a,b, c,d,ab,ac$ , which is impossible. Hence, one of the two numbers $d_6$ and $d_7$ is either a prime $p$ or the square $p^2$ of a prime $p\neq 2$ .

Let it be $d_i$ , $\left\{i, j\right\} = \left\{6, 7\right\}$ ; then $d_i | (d_j -1) (d_j +1)$ implies that $d_j\stackrel{d_i}{\equiv}\pm1$ , and consequently $(d_i^2 -1)/dj \stackrel{d_i}{\equiv}\pm1$ as well. But either $d_j$ or $(d_i^2 -1)/dj$ is less than $d_i$ , and therefore equals $d_i-1$ or equals $1$ . The only nontrivial possibilities are $(d_i^2 -1)/dj=1$ and $d_j = d_i \pm1$ . In the first case we get $d_i < d_j$ ; hence $d_7 =d_6^2 -1 = (d_6-1)(d_6+1)$ ; hence $d_6+1$ is a divisor of $n$ that is between $d_6$ and $d_7$ . This is impossible. We thus conclude that $d_7 = d_6 +1$ . Setting $d_6 = x$ , $d_7 = x+1$ we obtain that $n = x^2+(x+1)^2-1 = 2x(x+1)$ is even.

$(i)$ Assume that one of $x$ , $x+1$ is a prime $p$ . The other one has at most $6$ divisors and hence must be of the form $2^3, 2^4, 2^5, 2q, 2q^2, 4q$ , where $q$ is an odd prime. The numbers $2^3$ and $2^4$ are easily eliminated, while $2^5$ yields the solution $x = 31$ , $x+1 = 32$ , $n = 1984$ . Also, $2q$ is eliminated because $n = 4pq$ then has only $4$ divisors less than $x$ ; $2q^2$ is eliminated because $n=4pq^2$ has at least $6$ divisors less than $x$ ; $4q$ is also eliminated because $n= 8pq$ has $6$ divisors less than $x$ .

$(ii)$ Assume that one of $x$ , $x+1$ is $p^2$ . The other one has at most $5$ divisors ( $p$ excluded), and hence is of the form $2^3, 2^4, 2q$ , where $q$ is an odd prime. The number $2^3$ yields the solution $x = 8$ , $x+1 = 9$ , $n = 144$ , while $2^4$ is easily eliminated. Also, the number $2q$ is eliminated because $n = 4p^2q$ has $6$ divisors less than $x$ .

Thus there are two solutions in total: $144$ and $1984$ .