It is possible to solve the equation if $p = x^4 - 32x - 16$ is prime with quadratic reciprocity only. Maybe it is possible to generalize this proof.

There is a theorem of Gauss. Let $p \equiv 1 \pmod{8}$ be a prime, $p = a^2 + 8b^2$. Then 2 is a fourth power modulo $p$ if and only if $a \equiv \pm 1 \pmod{8}$. $p = x^4 - 32x - 16 = u^2 + 32v^2$ implies $u \equiv \pm 1 \pmod{8}$ and 2 is a fourth power modulo $p$. 

$$ p = (x^2 + 4)^2 - 2(2x + 4)^2$$
Modulo every odd prime divisor of $2x + 4$ we see that $p$ is a quadratic residue and by reciprocity $2x+4$ is a square modulo $p$. Similar analysis shows that whether $x^2 + 4$ is a square modulo $p$ depends on parity of the prime divisors of the forms $8k + 5$ and $8k + 7$. But $x^2 + 4 \equiv 5 \pmod{8}$ and therefore $x^2 + 4$ is not a square modulo $p$. Now modulo $p$ consideration shows that 2 is not a fourth power, which contradicts previous argument.