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Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Will Jagy showed that the representation $$ p = (x^2 + 4)^2 - 2(2x + 4)^2 $$ is primitive and $p$ is not divisible by any prime $q \equiv \pm 3 \pmod{8}$. Then from $p = u^2 + 32v^2$ we see that $d = \gcd(u, v) \equiv \pm 1 \pmod{8}$, $p = p_1d^2$, $p_1$ has only divisors of the form $8k + 1$. $u \equiv \pm 1 \pmod{8}$ since $p \equiv 1 \pmod{16}$. $$p_1 = \left(\frac{u}{d}\right)^2 + 32 \left(\frac{v}{d}\right)^2$$ Applying theorem 1 we obtain that $p$ is divisible by an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power.

$$p_1d^2 = (x^2 + 4)^2 - 2(2x + 4)^2$$ Applying theorem 2 we obtain $$\left(\frac{(x^2 + 4)(2x + 4)}{p_1}\right) = \left(\frac{(x^2 + 4)(2x + 4)}{p_1d^2}\right) = \left(\frac{-2}{x^2 + 4}\right) = -1 $$ Therefore $p$ is divisible by an odd number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power. And this is a contradiction.

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is $\pm 1$ modulo 8, $2 \mid \nu_q(p)$ for any prime $q \equiv 7 \pmod{8}$. Then $p$ has an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q^{\nu_q(p)}}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Will Jagy showed that the representation $$ p = (x^2 + 4)^2 - 2(2x + 4)^2 $$ is primitive and $p$ is not divisible by any prime $q \equiv \pm 3 \pmod{8}$. Then from $p = u^2 + 32v^2$ we see that $d = \gcd(u, v) \equiv \pm 1 \pmod{8}$, $p = p_1d^2$, $p_1$ has only divisors of the form $8k + 1$. $u \equiv \pm 1 \pmod{8}$ since $p \equiv 1 \pmod{16}$. $$p_1 = \left(\frac{u}{d}\right)^2 + 32 \left(\frac{v}{d}\right)^2$$ Applying theorem 1 we obtain that $p$ is divisible by an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power.

$$p_1d^2 = (x^2 + 4)^2 - 2(2x + 4)^2$$ Applying theorem 2 we obtain $$\left(\frac{(x^2 + 4)(2x + 4)}{p_1}\right) = \left(\frac{(x^2 + 4)(2x + 4)}{p_1d^2}\right) = \left(\frac{-2}{x^2 + 4}\right) = -1 $$ Therefore $p$ is divisible by an odd number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power. And this is a contradiction.

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Will Jagy showed that the representation $$ p = (x^2 + 4)^2 - 2(2x + 4)^2 $$ is primitive and $p$ is not divisible by any prime $q \equiv \pm 3 \pmod{8}$. Then from $p = u^2 + 32v^2$ we see that $d = \gcd(u, v) \equiv \pm 1 \pmod{8}$, $p = p_1d^2$, $p_1$ has only divisors of the form $8k + 1$. $u \equiv \pm 1 \pmod{8}$ since $p \equiv 1 \pmod{16}$. $$p_1 = \left(\frac{u}{d}\right)^2 + 8 \left(\frac{v}{d}\right)^2$$ Applying theorem 1 we obtain that $p$ is divisible by an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power.

$$p_1d^2 = (x^2 + 4)^2 - 2(2x + 4)^2$$ Applying theorem 2 we obtain $$\left(\frac{(x^2 + 4)(2x + 4)}{p_1}\right) = \left(\frac{(x^2 + 4)(2x + 4)}{p_1d^2}\right) = \left(\frac{-2}{x^2 + 4}\right) = -1 $$ Therefore $p$ is divisible by an odd number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power. And this is a contradiction.

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Will Jagy showed that the representation $$ p = (x^2 + 4)^2 - 2(2x + 4)^2 $$ is primitive and $p$ is not divisible by any prime $q \equiv \pm 3 \pmod{8}$. Then from $p = u^2 + 32v^2$ we see that $d = \gcd(u, v) \equiv \pm 1 \pmod{8}$, $p = p_1d^2$, $p_1$ has only divisors of the form $8k + 1$. $u \equiv \pm 1 \pmod{8}$ since $p \equiv 1 \pmod{16}$. $$p_1 = \left(\frac{u}{d}\right)^2 + 32 \left(\frac{v}{d}\right)^2$$ Applying theorem 1 we obtain that $p$ is divisible by an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power.

$$p_1d^2 = (x^2 + 4)^2 - 2(2x + 4)^2$$ Applying theorem 2 we obtain $$\left(\frac{(x^2 + 4)(2x + 4)}{p_1}\right) = \left(\frac{(x^2 + 4)(2x + 4)}{p_1d^2}\right) = \left(\frac{-2}{x^2 + 4}\right) = -1 $$ Therefore $p$ is divisible by an odd number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power. And this is a contradiction.

It is possible to solve the equation if $p = x^4 - 32x - 16$ is prime with quadratic reciprocity only. Update: Now the unsolvability proof should work for composite numbers as well.

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.

This proof can be generalized to the case of composite $p$ using the following theorems:

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Update 2. Now the proof should be more readable. I deleted some content because it is replaced by more elegant version. The equation is unsolvable. The proof requires two theorems

Theorem 1. Let $p = a^2 + 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv \pm 1 \pmod{8}$.

Theorem 2. Let $p = a^2 - 8b^2 \equiv 1 \pmod{8}$, $\gcd(a, b) = 1$, every prime divisor of $p$ is 1 modulo 8. Then $p$ has an even number of prime divisors for which 2 is not a fourth power if and only if $a \equiv 1, 3 \pmod{8}$.

I will prove the first theorem. Let $q$ be a prime divisor of $p$. $-(ab)^2 \equiv 8b^4 \pmod{q}$. Since -1 is a fourth power modulo $q$, 2 will be a fourth power modulo $q$ if and only if $ab$ is a quadratic residue. Therefore we can reformulate the theorem in terms of Jacobi symbols: $$\prod_{q \mid p, q \in \mathbb{P}} \left(\frac{ab}{q}\right) = \left(\frac{ab}{p}\right) = \left(\frac{2}{a}\right)$$ And the proof becomes simple. $$\left(\frac{ab}{p}\right) = \left(\frac{a}{p}\right)\left(\frac{b_1}{p}\right) = \left(\frac{a^2 + 8b^2}{a}\right)\left(\frac{a^2 + 8b^2}{b_1}\right) = \left(\frac{2}{a}\right)$$ where $b = b_1 \cdot 2^k$, $b_1 \equiv 1 \pmod{2}$.

Will Jagy showed that the representation $$ p = (x^2 + 4)^2 - 2(2x + 4)^2 $$ is primitive and $p$ is not divisible by any prime $q \equiv \pm 3 \pmod{8}$. Then from $p = u^2 + 32v^2$ we see that $d = \gcd(u, v) \equiv \pm 1 \pmod{8}$, $p = p_1d^2$, $p_1$ has only divisors of the form $8k + 1$. $u \equiv \pm 1 \pmod{8}$ since $p \equiv 1 \pmod{16}$. $$p_1 = \left(\frac{u}{d}\right)^2 + 8 \left(\frac{v}{d}\right)^2$$ Applying theorem 1 we obtain that $p$ is divisible by an even number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power.

$$p_1d^2 = (x^2 + 4)^2 - 2(2x + 4)^2$$ Applying theorem 2 we obtain $$\left(\frac{(x^2 + 4)(2x + 4)}{p_1}\right) = \left(\frac{(x^2 + 4)(2x + 4)}{p_1d^2}\right) = \left(\frac{-2}{x^2 + 4}\right) = -1 $$ Therefore $p$ is divisible by an odd number of prime divisors of the form $8k + 1$ for which 2 is not a fourth power. And this is a contradiction.