Selection Examination

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 $$2^x\cdot 1009^x=2^{2y}\cdot 5^{2y}+2^z\cdot 7^z\cdot 137^z,(1)$$

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 $$2^z.(3)\text{From (1) and (3) we get }x=z.$$ Then the equation has the form: $$2018^z-1918^z=100^y\iff 2^z(1009^z-7^z\cdot 137^z)=2^{2y}\cdot 5^{2y}.(4)$$ 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 $2z\le z+1\iff z\le 1\iff z=1$. It follows that the only solution is $(x,y,z)=(1,1,1)$.