Find integer triples $(x,y,z)$ such that $2018^x=y^2+z^2+1$ 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?

You have solved almost all cases. So let me add the solution for $x=1$, and exclude negative $x$.
Since $2017$ is a prime which is congruent $1$ modulo $4$, it is representable as the sum of two squares. The solutions are, up to sign and order, given by
$$
2017=44^2+9^2,
$$
see here:
$2017$ as the sum of two squares
