This is really trivial, CLearly as you can see all of $x,y,z$(assumed $\in\mathbb{N}$) is even and suppose they have some odd divisors say $x=2^a.p,y=2^b.q,z=2^c.r $ where $p,q,r\geq 1$ Now, Let, $a=\text {min} \{a,b,c\}$ $2^{2a}(p^2+2^{2b-2a}q^2+2^{2c-2a}r^2)=2^{2004}$ but this a contradiction since LHS would atleast have an odd prime divisor. Forcing , $q=0,r=0,p=1$ and $2^{2a}=2^{2004}$ So only possible solution is $(2^{1002},0,0)$ or its permutations; And If $\in\mathbb{Z}$ then $-2^{1002}$ also be a solution;

Konigsberg wrote: Find all triples $(x,y,z)$ such that $x^2+y^2+z^2=2^{2004}$. $x^2+y^2+z^2=2^{2004}$.Clearly among at least 2 no.s may be odd or three of them is even.If 2 of them is odd then LHS must have a odd prime factor but RHS don't have.so three of them must be even. Let, $x=2x_1,y=2y_1,z=2z_1$. so, $x^2_1 +y^2_1 +z^2_1=2^{2000}$.Similarly as above after doing it $1002$ times we get, $x^2_{1002} +y^2_{1002} +z^2_{1002}=1$.so $(x_{1002},y_{1002},z_{1002})=(\pm1,0,0) \implies (x,y,z)=(\pm2^{1002},0,0)$ and all its permutations.