The only solution is $n=500$. Claim: $n$ is even. Proof: Assume that $n$ is odd, then all of its divisors are odd and $d_5^2+d_6^2$ is even, wich is not true. $\blacksquare$ Claim: $d_5$ and $d_6$ have the same prime divisors and both of them have less than $2$ prime divisors. Proof: Note that $n=d_5d_{k-4}=d_6d_{k-5}$ and $n=d_5d_{k-4}=d_5^2+d_6^2\Longleftrightarrow d_{k-4}-d_5=\frac{d_6^2}{d_5}$ wich shows that prime divisors of $d_5$ also divide $d_6$ and in the same way we show that prime divisors of $d_6$ divide $d_5$. If they have more than $3$ prime divisors then let $p_1,p_2,p_3$ be the least prime divisors of $n$ then $n$ is also divisible by $1,p_1,p_2,p_3,p_1p_2,p_1p_3,p_2p_3$ are less than $d_5,d_6$ wich is not true. $\blacksquare$ Claim: $d_5$ and $d_6$ are even. Proof: If not, then $p,q>2$ are divisors of $d_5,d_6$ then $1,2,p,q,2p,2q$ are divisors of $n$ less than $d_5,d_6$ wich is not true. $\blacksquare$ Therefore $n$ is divisible by $4$ and now we have three cases: If first divisors of $n$ are $1,2,3,4,6,12$, then $n=12^2+6^2$ implies $5\mid n$ wich is not true. If first divisors of $n$ are $1,2,4,p,2p,4p$ then $n=4p^2+16p^2=20p^2$ wich implies that $p=5$ and gives $n=2^25^3=500$ a solution. If first $6$ divisors of $n$ are $1,2,2^2,\dots 2^5$ then $n=2^8\cdot 5$ wich is not a solution.