This is solvable if and only if $p=2$ or $p\equiv 1\pmod{8}$. Note that for $p=2$, $x$ odd works. Now, we show $x^4\equiv -1\pmod{p}$ implies $8\mid p-1$. Let $d$ be the smallest positive integer such that $x^d\equiv 1\pmod{p}$. Then, it is well-known that for any $k$ with $x^k\equiv 1\pmod{p}$, $d\mid k$. Now, $x^4\equiv -1\pmod{p}$ and $x^8\equiv 1\pmod{p}$ implies $d\nmid 4$, and $d\mid 8$, that is $d=8$. Moreover, Fermat's theorem also tells us that $x^{p-1}\equiv 1\pmod{p}$, and thus, $8\mid p-1$. We now show $p\equiv 1\pmod{8}$ works. Let $g$ be a primitive root modulo $p$, and set $x=g^{\frac{p-1}{8}}$. Note that, $x^4 =g^{\frac{p-1}{2}}$. Now, by Fermat's theorem, $x^8\equiv 1\pmod{p}$, and thus $x^4\equiv \pm 1\pmod{p}$. But since $g$ is a primitive root, $x^4\equiv 1\pmod{p}$ is impossible. Hence, $x^4+1 \equiv 0\pmod{p}$.