Recall that the ring of Gaussian integers is $$\mathbb Z[i]=\{a+bi:\ a,b\in\mathbb Z\}.$$ Clearly $$(a+bi)^2=a^2-b^2+2abi\ \ \mbox{and}\ \ (a+bi)^4=(a^2-b^2)^2-4a^2b^2+4ab(a^2-b^2)i.$$
Question. Is it true that $\{x^4+y^2+z^2:\ x,y,z\in\mathbb Z[i]\}=\{a+2bi:\ a,b\in\mathbb Z\}$?
Evidence. Via Mathematica I have found that \begin{align} &\{x^4+y^2+z^2:\ x,y,z\in\{r+si:\ r,s\in\{-14,\ldots,14\}\}\} \\&\quad \supseteq\{a+2bi:\ a,b\in\mathbb Z\ \mbox{and}\ |a+2bi|\le 50\}. \end{align} For example, $$43+22i=2^4+(14-11i)^2+(11-13i)^2$$ and $$-34+26i=(2+i)^4+(13-i)^2+(1+14i)^2=(4+i)^4+(11-11i)^2+(1+14i)^2.$$
Motivation. The question is motivated by my following result (cf. my 2017 JNT paper) $$\{x^4+y^2+z^2+w^2:\ x,y,z,w=0,1,2,\ldots\}=\{0,1,2,\ldots\}$$ which refines Lagrange's four-square theorem.
I conjecture that the question has a positive answer, but I'm unable to prove this. Your comments are welcome!