We divide into cases. Case 1: $ i^x = (1 + i)^y \neq 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 \neq (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 \neq 1 $ is satisfied as long as $ y \neq 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 \neq 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.