If z≥2 x^2+y^4 is 3 mod 4, which is not possible, hence the only nontrivial solution is (2,1,1) and its variations

I think $(0,0,0)$ is also a possibility other than$(2,1,1)$

$(0,0,0)$, $(2,1,1)$ and $(-2,-1,1)$ are the only solutions to the diophantine.

Actually you guys are forgetting negative solutions, $(-2,-1,1)$ also works

From a cursory glance , we see that $(0,0,0)$ in an eligible solution. We consider the equation $\mod {4}$. For $z=1$ we get $RHS\equiv2(mod 4)$ , and in $LHS$ we have possible remainders for $x^2 \equiv0,1,2(mod 4)$ , $y^4\equiv0,1(mod 4)$ and therefore we can have $x^2+y^4+1\equiv2(mod4)$ , hence no contradiction ; thus we have solutions for $z=1$ ; so the only pairs that satistfy this relation are $(2,1,1)$ and $(-2,-1,1)$ For $z \ge 2$ , we get $LHS \equiv 0$ , and then we observe that $x^2+y^4+1\not\equiv0(mod4)$ , thus a contradiction , and hence no solutions for this case. Therefore the only solutions in $\mathbb{Z}$ are $(0,0,0) , (2,1,1) , (-2,-1,1) , (-2,1,1) , (2,-1,1) $.

how is (0,0,0) a solution x or y power to 0 gives 1 people

@^ $(0,0,0)$ works recursively because if you plug in the values into the equation $x^2+y^4+1=6^z$, it eventually leads to something like this: Assuming that $x=y=z=0$, then the equation can be simplified to $(0^2+0^4+1)=6^0$. Let's us not forget that any real number taken to the power of $0$ is equivalent to one. Therefore, we can rewrite the equation as $(0+0+1)=1$ which of course is true. Therefore a solution for $\mathbb{Z}$ is $(0,0,0)$.

@^ yeah of course what was i thinking i assumed that all the zeros are the powers

There are also the two "mixed" solutions $(-2,1,1)$ and $(2,-1,1)$.