We write $2018 = 2 \cdot 1009$, $100 = 2^2 \cdot 5^2$ and $1918 = 2 \cdot 7 \cdot 137$. Then, the equation has the form
$$2^x \cdot 1009^x = 2^{2y} \cdot 5^{2y} + 2^z \cdot 7^z \cdot 137^z, \quad (1)$$

Consider the following cases:

* If $2y < z$, then the power of $2$ that divides the right hand side is $2^{2y}$. (2)
From (1) and (2) we get that $x = 2y$. The equation takes the form:
$$1918^z = 2018^{2y} - 100^{y} = (2018^y - 10^y)(2018^y + 10^y).$$
However, $2010 \mid 2018^y - 10^y$, so $2010 \mid 1918^z$, a contradiction.

* If $z < 2y$, then the highest power of $2$ that divides the right hand side is
$$2^z. \quad (3) \quad \text{From (1) and (3) we get } x = z.$$
Then the equation has the form:
$$2018^z - 1918^z = 100^y \Leftrightarrow 2^z (1009^z - 7^z \cdot 137^z) = 2^{2y} \cdot 5^{2y}. \quad (4)$$
Since $z$ is odd, the number $1009^z - 7^z \cdot 137^z$ is even but it is not divisible by $4$, so the highest power of $2$ that divides the left hand side is $z+1$, so $2y = z+1$, and the equation takes the form $2018^z - 1918^z = 10^{z+1}$. Therefore, $10^{z+1} = 2018^z - 1918^z \ge 100^z = 10^{2z}$, so $2z \le z+1 \Leftrightarrow z \le 1 \Leftrightarrow z = 1$.
It follows that the only solution is $(x, y, z) = (1, 1, 1)$.