To find the triples, I had some tries. First , consider $x\ge 2$, then $2018^x=0\pmod 4$. But for the right side $y^2+z^2+1$. Consider 3 situations.
- case 1. Both $y$ and $z$ are even. Then $y^2+z^2+1=1\pmod 4$.
- case 2. One of them are even, the other is odd. Then $y^2+z^2+1=2\pmod 4$.
- case 3. Both $y$ and $z$ are odd. Then $y^2+z^2+1=3\pmod 4$.
Thus for $x\ge 2$, there is no solution. Then $x=0$ or $x=1$.
Now for $x=0$, $2018^0=y^2+z^2+1$ ,then $y^2+z^2=0$, hence we have $y=0$ and $z=0$.
For $x=1$, $2018^1=y^2+z^2+1$, then my question is how to solve $y^2+z^2=2017$.
Also since I didn't learn Number Theory systematically, I don't know if I was proceeding in the right way for this question.
Lastly, Are there more general methods for problems like this?