Case 1: $n$ is a prime number $\Rightarrow LHS = 1.$ Case 2: $n$ is the product of two primes $p,q$. $$n = pq \Rightarrow LHS = 1+p^2+q^2 = 5(n+1) = 5(pq+1)$$$$\Leftrightarrow p^2-5pq+q^2-4 = 0 \Rightarrow p_{1,2} = \frac{5q \pm \sqrt{25q^2-4q^2+16}}{2}$$$$\Rightarrow 25q^2-4q^2+16 = k^2 \Leftrightarrow 21q^2+16 = k^2$$$$\Leftrightarrow 21q^2 = (k-4)(k+4)$$The rest is casework. Case 3: $n$ has at least $3$ prime factors ($p,q$ primes, $r = \frac{n}{pq} > 1$). $$n = pqr \Rightarrow LHS \geqslant 1+p^2+q^2+r^2+p^2q^2+q^2r^2+r^2p^2$$$$\overset{AM-GM}{\geqslant} 1+3(pqr)^\frac{2}{3}+3(pqr)^\frac{4}{3} = 1+3n^\frac{2}{3}+3n^\frac{4}{3} > 5(n+1)\quad \forall n\geqslant 3$$