Keep on Reducing

${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=8xyz$

On the left side of the first equation, exactly one term is even or all three terms are even. If exactly one term is even, then the right side is divisible by $4$ , the left only by $2$ . A contradiction. Hence all three terms are even, so let say $x={ 2x }_{ 1 },y={ 2y }_{ 1 },z={ 2z }_{ 1 }$ . Applying this in the first equation we get:

${ x }_{ 1 }^{ 2 }+{ y }_{ 1 }^{ 2 }+{ z }_{ 1 }^{ 2 }=16{ x }_{ 1 }{ y }_{ 1 }{ z }_{ 1 }$ .

From this, with the same reasoning we get ${ x }_{ 1 }=2{ x }_{ 2 },{ y }_{ 1 }={ 2y }_{ 2 },z_{ 1 }={ 2z }_{ 2 }$ , and

${ x }_{ 2 }^{ 2 }+{ y }_{ 2 }^{ 2 }+{ z }_{ 2 }^{ 2 }=32{ x }_{ 2 }{ y }_{ 2 }{ z }_{ 2 }$ .

Again, it follows that $x_{ 2 },{ y }_{ 2 },{ z }_{ 2 }$ are even, and so on, that is

$x=2x_{ 1 }={ 4x }_{ 2 }={ 8x }_{ 3 }=...={ 2 }^{ n }{ x }_{ n }$

$y=2y_{ 1 }={ 4y }_{ 2 }={ 8y }_{ 3 }=...={ 2 }^{ n }{ y }_{ n }$

$z=2z_{ 1 }={ 4z }_{ 2 }={ 8z }_{ 3 }=...={ 2 }^{ n }{ z }_{ n }$

so, as we see $x,y,z$ are divisible by ${ 2 }^{ n }$ for any $n$ . This is only possible for $x=y=z=0$ , but the instructions make exception of this answer, so the answer is $\boxed { 0 }$ .

Also, it is well known that the equation ${ x }^{ 2 }+{ y }^{ 2 }+{ z }^{ 2 }=kxyz$ has only solution for $k=1$ and $k=3$ infinitely many solutions.

Subtracting $2yz$ from both sides, we have $x^2+(y-z)^2=2yz(4x-1)$ . Since $(y-z)^2 \geq 0$ , we have $x^2-(8yz) x+2yz \leq 0$ . Since $x$ is real, the discriminant of this quadratic in $x$ must be non-positive. This gives $yz(yz-\frac{1}{8})\leq 0$ , i.e. $0\leq yz \leq \frac{1}{8}$ , which is impossible to hold for integral non-zero $y,z$ . Also if one of them is zero, the only real solution of the given equation is the all-zero solution. Thus no solution excepting from all-zero.