$|i^x|=1$ always, while $(1+i)^y$ has magnitude $2^{\frac{y}{2}}$

Split into three cases by which two of the three entries coincide, keeping the third distinct

We need exactly two of $i^x,$ $(1+i)^y,$ $z$ equal, with the third different. The three cases are the three possible equal pairs.

Case $i^x=(1+i)^y:$ since $|i^x|=1$ but $|(1+i)^y|=2^{\frac{y}{2}}\gt1$ for $y\ge1,$ we need $y=0,$ so $(1+i)^0=1$ and $i^x=1,$ i.e. $x\in\{0,4,8,12,16\}.$ Then $z$ is any of the $19$ values other than $1.$ This gives $5\cdot19=95$ triples.

Case $i^x=z:$ the only nonnegative-integer value of $i^x$ is $1$ (with $x$ a multiple of $4$), so $z=1$ and $(1+i)^y\ne1,$ meaning $y\ge1.$ This gives $5\cdot19=95$ triples.

Case $(1+i)^y=z:$ since $(1+i)^2=2i,$ the power $(1+i)^y$ is a nonnegative integer below $20$ only for $y=0$ (value $1$) or $y=8$ (value $16$). If $y=0,\ z=1,$ we need $i^x\ne1,$ so $x$ is not a multiple of $4$ ($15$ values). If $y=8,\ z=16,$ then $i^x$ is never $16,$ so $x$ is free ($20$ values). This gives $15+20=35$ triples.

Altogether $95+95+35=225.$

Thus, the correct answer is D.