We claim that there is no such $n$. Let $1,d_2,d_3, d_4, d_5$ be the $5$ smallest divisors of $n$ in increasing order. $x^2=1+d_2^2+d_3^2+d_4^2+d_5^2$ If $n$ is odd, then all $d_i$ are odd and hence, $x^2=RHS\equiv 5\pmod 8$, not possible. So, $d_2=2$ If $4|n$, then we take two cases If $3|n$, $d_3=3,d_4=4$ gives $x^2=30+d_5^2$ but this equation has no solution(just see mod 4) If $3\nmid n$, then $d_3=4$. Both $d_4$ and $d_5$ are even due to mod 4. But if $d_4=2^ab$ where $b$ is odd, then $b|n$ and $b<d_4$, contradiction. So, $d_4,d_5$ are powers of $2$ and hence, $x^2\equiv 3\pmod 4$, not possible. So, $4\nmid n$ and hence $d_3$ is an odd. Both $d_4$ and $d_5$ must be odd, by $\mod 4$. If $d_3\geq 5$, $x^2=5+d_3^2+d_4^2+d_5^2\equiv 2\pmod 6$, not possible. So, $d_3=3$ But, $d_3=3\implies 6|n$ so one of $d_4,d_5$ is $6$, contradiction to the fact that $d_4,d_5$ are odd. Hence, we have exhausted all cases and conclude that no such $n$ exists.

A shorter proof might be possible using $\mod 12$, but this seems a bit natural