Notice how nicely factorizes $x^4+y^4+z^4-2x^2y^2-2y^2z^2-2z^2x^2=(x^2+y^2-z^2)^2 - (2xy)^2=$ $((x+y)^2-z^2)((x-y)^2 - z^2) = $ $(x+y+z)(x+y-z)(x-y+z)(x-y-z)$.

That still does not answer the question.

It's proved by reducing in modular arithmetic: $x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2 \equiv 2 \pmod 3$ (I used the fact $x^4\equiv x^2 \pmod 3$ by the way), the squares are 0 or 1 so we have to check 8 cases (4 by symmetry). * $(x^2,y^2,z^2)=(0,0,0)$ then $0 \not \equiv 2 \pmod 3$ not a solution * $(x^2,y^2,z^2)=(1,0,0)$ then $1 \not \equiv 2 \pmod 3$ not a solution * $(x^2,y^2,z^2)=(1,1,0)$ then $0 \not \equiv 2 \pmod 3$ not a solution * $(x^2,y^2,z^2)=(1,1,1)$ then $0 \not \equiv 2 \pmod 3$ not a solution Therefore no solutions.

I think this is what mavropnevma wanted to suggest. Consider \begin{align*} S&:=x^4+y^4+z^4-2y^2z^2-2z^2x^2-2x^2y^2 \\ &=-(x+y+z)(y+z-x)(z+x-y)(x+y-z)\,. \end{align*} All calculations are done in $\mathbb{F}_3$. Suppose that $S=2000 = 2$. Since none of the four factors is divisible by $3$, among $y+z-x$, $z+x-y$, and $x+y-z$, we must have two which have the same residue modulo $3$. WLOG, $y+z-x=z+x-y$. Thus, $x=y$. Now, we get \[S=-z^2(2x+z)(2x-z)=z^2(z+2x)(z-2x)\,.\] If $x \neq 0$, then $z-2x$, $z$, or $z+2x$ must be divisible by $3$, a contradiction. Therefore, $x=0$, but then $S=z^4 \in \{0,1\}$, another contradiction.

Actually, the factorization mavropnevma did, does lead to other simpler solution. Notice that all of the factors are congruent modulo 2. Since 2000 has a 2 raised to the fourth power, all of the factors MUST have a factor 2 and a factor 2 only. Checking the possibilities modulo 4 we find that in all cases at least one of them is also divisible by 4, contradiction.