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,
 A: The purpose of this answer is to give another proof of the generalization suggested in lhf's answer:

Let $p$ be an odd prime, and let $k$ be a positive integer.  Suppose there is $x \in \mathbb{Z}$ such that $x^{2^k} \equiv -1 \pmod p$.  Then $p \equiv 1 \pmod{2^{k+1}}$.

Proof: We work in the unit group $U(p) = (\mathbb{Z}/p\mathbb{Z})^{\times}$.  Since
$x^{2^k} \equiv -1$, $x^{2^{k+1}} \equiv (x^{2^k})^2 \equiv (-1)^2 \equiv 1 \pmod p$.  Thus the order of $x$ in $U(p)$ divides $2^{k+1}$.  To show that it is not smaller, we recall the following

Fact: Let $G$ be a group, $x$ an element of $G$ of finite order $n$, and $a \in \mathbb{Z}^+$.  Then the order of $x^a$ is $\frac{n}{\operatorname{gcd}(n,a)}$.

Proof of the Fact: It is no loss of generality to assume that $x$ generates $G$ and thus that $G \cong (\mathbb{Z}/n\mathbb{Z},+)$.  For the (easy) proof in this special case, see e.g. Proposition 7 of these notes.
Returning to the main proof, we know that the order of $x$ in $U(p)$ is of the form $n = 2^l$ for some $l \leq k+1$.  Since $x^{2^k} = -1$, $x^{2^k}$ has order $2$.  Applying the fact with $G = U(p)$, $a = 2^k$ gives
$2 = \frac{2^l}{\operatorname{gcd}(2^l,2^k)} = 2^{l - \operatorname{min}(k,l)}$,
so
$l - \operatorname{min}(k,l) = 1$
and thus $l = k+1$.
A: As you have noticed, $p \equiv 1 \bmod 4$. Then $1 \equiv n^{p-1} = (n^4)^{(p-1)/4} \equiv (-1)^{(p-1)/4} \bmod p$. This means that $(p-1)/4$ is even, i.e., $p\equiv 1 \bmod 8$.
By induction, this argument generalizes to: if an odd prime $p$ divides a number of the form $n^k+1$, where $k$ is a power of $2$, then $p \equiv 1 \bmod {2k}$.
A: 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.
