Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$ Problem 

Prove that if $p$ is an odd prime that divides a number of the form $n^4 + 1$ then $p \equiv 1 \pmod{8}$

My attempt was,
Since $p$ divides $n^4 + 1 \implies n^4 + 1 \equiv 0 \pmod{p} \Leftrightarrow n^4 \equiv -1 \pmod{p}$.
It follows that $(n^2)^2 \equiv -1 \pmod{p}$, which implies $-1$ is quadratic residue modulo $p$. Hence $p \equiv 1 \pmod{4} \Leftrightarrow p \equiv 1 \pmod{8}$.
Am I in the right track? 
Thanks,

If $$p \mid (n^k+1),  $$
$$n^k \equiv -1 \pmod{p}$$
$$n^{2k} \equiv 1 \pmod{p}$$
If$$ \operatorname{ord}_pn=d,$$ then $d \mid 2k$.
If $d \mid k$, then  $$n^k\equiv 1 \pmod{p}$$ $\Rightarrow$ $$-1\equiv 1 \pmod{p}$$ 
$\Rightarrow p\mid 2$ which is impossible as $p$ is odd prime $\Rightarrow d\nmid k$.
If $(k,2)=1$ i.e., $k$ is odd, $d$ can divide $2$ $\Rightarrow$ $d=2$ as $d \nmid k \Rightarrow d \neq 1$.
In that case,$$ p \mid (n^2-1) \text{, or } p \mid (n+1) \text{ as } d\neq 1.$$ 
Then $d$ will be $2k$ if $d \neq 2$ i.e., iff $p \nmid (n+1)$.
If $k$ is $2^r$ where integer r ≥1, then $d \nmid 2$ as $d \nmid k$, then $d=2k$.
But $d \mid (p-1) \Rightarrow p≡1 \pmod{2k}$ if $k$  is of the form $2^r$ where integer r ≥1.
