My approach:
For $p=2$, we have $(x+1)^4=x^4+1$ so reducible, let's assume $p>2$.
Suppose $F$ is finite, if $|F|=p^r$ with $r\geq2$, then we can find an element $a\in F$ such that $a^4=-1$; if $|F|=p^r$ with $r=1$, let $a$ be the root of $f$, by contradiction we have $[F_{p^2}:F(a)][F(a):F]=2$ ($F_{p^2}$ an extension field of $F$ with order $p^2$) and $[F(a):F]=4$ which is impossible.
Suppose $F$ is infinite. if the below statement is true then the result follows from finite case
For any infinite field $F$ with characteristic $p>0$, it contains finite subfields with order $p^r$ for all $r>0$
I have read a post somewhere that the above statement is true for $r=1$, but I don't know if it is still true for other $r$.
If $\sqrt{-1} \in F$ then $x^4+1 = \left(x^2+\sqrt{-1}\right)\left(x^2-\sqrt{-1}\right)$.
If $\sqrt{2} \in F$ then $x^4+1 = \left(x^2+\sqrt{2}x+1\right)\left(x^2-\sqrt{2}x+1\right)$.
If $\sqrt{-2} \in F$ then $x^4+1 = \left(x^2+\sqrt{-2}x-1\right)\left(x^2-\sqrt{-2}x-1\right)$.
Since $\text{char }F = p$, we have the subfield $\mathbb{F}_p$ generated by $1$.
If $-1$ is a square mod $p$ we are done since $\sqrt{-1} \in \mathbb{F}_p \subseteq F$.
Otherwise, $\left(\frac{-2}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{2}{p}\right) = -\left(\frac{2}{p}\right)$, so if $p$ is odd then either $2$ or $-2$ is a square mod $p$.
If $p = 2$ then $-1$ is a square mod $p$ so in all cases we are done and $x^4+1$ is reducible in $F$ (in fact, over the subfield $\mathbb{F}_p$).