jmc

We divide into cases.

Case 1: $i^x=(1+i{)}^y\ne z.$

Note that $|i^x|=|i{|}^x=1$ and $|(1+i{)}^y|=|1+i{|}^y=(\sqrt{2}{)}^y,$ so we must have $y=0.$ Then $i^x=1$ only when $x$ is a multiple of 4. There are 5 possible values of $x$ (0, 4, 8, 12, 16), and 19 possible values of $z,$ so there are $5\cdot 19=95$ triples in this case.

Case 2: $i^x=z\ne (1+i{)}^y.$

The only way that $i^x$ can be a nonnegative integer is if it is equal to 1, which in turn means that $x$ is a multiple of 4. As in case 1, $|(1+i{)}^y|=(\sqrt{2}{)}^y,$ so $(1+i{)}^y\ne 1$ is satisfied as long as $y\ne 0.$ This gives us 5 possible values of $x,$ and 19 possible values of $y,$ so there are $5\cdot 19=95$ triples in this case.

Case 3: $(1+i{)}^y=z\ne i^x.$

Note that $(1+i{)}^2=2i,$ and we must raise $2i$ to a fourth power to get a nonnegative integer. Hence, $(1+i{)}^y$ is a nonnegative integer only when $y$ is a multiple if 8. Furthermore, $(1+i{)}^8=(2i{)}^4=16,$ and $(1+i{)}^{16}=16^2=256,$ so the only possible values of $y$ are 0 and 8.

For $y=0,$ $z=1,$ and then $x$ cannot be a multiple of 4. This gives us $20-5=15$ triples.

For $y=8,$ $z=16,$ and $x$ can take on any value. This gives us 20 triples, so there are $15+20=35$ triples in this case.

Therefore, there are a total of $95+95+35=\boxed{225}$ triples.