The equation $(1)$ has not integer solution. An elementary proof.
Suppose $(x,y,z)=(a,b,c)$ is a solution and consider as unknowns $(X,Y)=(4,2)$ so we can form the system $$\begin{cases}b^2X+(a+c^2)Y=-(1+a^2b)\\-X+7Y=10\end{cases}$$ whose solution is $$X=\dfrac{-7-7a^2b-10a-10c^2}{7b^2-a-c^2}$$ $$Y=\dfrac{10b^2+1+a^2b}{7b^2-a-c^2}$$ Because of $X=2Y$, one has after simplification $$9(1+a^2b)+10(a+c^2+2b^2)=0$$ and multiplying equation $1+2a+a^2b+4b^2+2c^2=0$ by $5$ we have $$5(1+a^2b)+10(a+c^2+2b^2)=0$$ Subtraction now gives $$4(1+a^2b)=0$$ We are done.