Solution:

Expand and rewrite: $$x(x-y)+y(y-z)+z(z-x)=\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}.$$

For the value to be $1$, the three nonnegative square terms must sum to $2$. Since $x,y,z$ are integers, this means the squared differences are $1,1,0$.

Thus two of the variables must be equal, and the third must differ from them by $1$. The condition $x=z$ and $y-1=x$ guarantees exactly that.

Thus, the answer is D .