Putting $z=1-x-y$ into the second equation gives $x^5+y^5+(1-x-y)^5-1=0$ which factors: $-5(x-1)(y-1)(x+y)(1-x+x^2-y+xy+y^2)=0$ It is easy to solve this equation to get: $x=1$ or $y=1$ or $y=-x$ or $y=\frac{1-x\pm\sqrt{-3+2x-3x^2}}{2}$ The solution with the square-root can be discarded since $-3+2x-3x^2<0$ for all $x\in\mathbb{R}$. By symmetry, we could have eliminated $x$ or $y$ instead. So the full solution set for $(x,y,z)$ is: $(1,t,-t)$, $(-t,1,t)$, $(t,-t,1)$ where $t\in\mathbb{R}$.