Given that $n > 3$ and $z$ is a Gaussian integer, when can $z^n \pm z$ be a rational integer? I came across the following conjecture. If you have any thoughts on how to approach it, let me know.
Conjecture. For any integer $n > 3$ and any Gaussian integer $z$ that is not a unit, if $z^n - z$ or $z^n + z$ is a rational integer, then $z$ is a rational integer.
 A: Your conjecture is likely to be true. I was not able to check the details because I couldn't get all the references, but here are the main steps.
The main point is that if there is a Gaussian integer $z$ such that $z^n-z-a=0$ or $z^n+z-a=0$ for some $a\in\mathbb{Z}$, and if $z$ is not an integer ,then $X^n-X-a$ or $X^n+X-a$ has a quadratic monic irreducible factor, namely the minimal polynomial of $z$.
Then, it is enough to understand which quadratic factors may occur in polynomials on the form $X^n\pm X-a$, and check that in each case they are not of the form $X^2-2u X+(u^2+v^2)$ (the minimal polynomial of $u+iv$
when $v\neq0$).
This question seems to be well understood.
In the paper [1] S. Rabinowitz, The Factorization of $x^5\pm x+n$ Math. Mag., 61 (1988), 191–193,
Rabinowitz proved that if $x^n\pm x+n, n\in\mathbb{Z}$ has an irreducible quadratic factor, then $n\in\{\pm 1,\pm 6,\pm 15,\pm 22440,\pm 2959640\}$.
The explicit factorisations are given in the paper (easy to find on the web) , and none of the quadratic factors have the right shape, so this is OK.
For $n>6$, Le proved that if $X^n-X-a$ has
has an irreducible quadratic factor $g$ which is monic,then either $n≡2(mod 6),a=-1$ and $g(x)=x^2-x+1,$ or $n=7,a=±280$ and $g=x^2\pm x+5.$
Once again, this is OK in this case.
Source: [2] M. H. Le, Irreducible quadratic factors of the trinomial $x^n-x-a$ J. Math.
(Wuhan), 24 (2004), 635–637.
I couldn't get the paper, and I have some doubts on the assumption $n>6$. It is maybe $n\geq 6$, i don't know.(ANyway, I think one can manage the case $n= 6$ by hand...)
Apparently, there are similar results for $X^n+X-a$, but I couldn't get the paper nor see the statements of the results.
Source: [3] M. Y. Lin, The irreducible quadratic factor of the trinomial $x^n+x-a$, Math. Appl. (Wuhan), 19 (2006), 656–658.
All in all, I think you conjecture should be true.
If you are able to get a copy of papers [2] and [3], you should be able to check it.
A: The question can be rephrased: Find the integers $n > 3$ for which there exist Gaussian integers $z$ such that
$z^n \pm z = \overline{z}^n \pm \overline{z}$. Rewriting the equation as $(z^n -\overline{z}^n)/(z-\overline{z}) = \mp 1$, one recognizes an instance of the problem of finding the  terms of a Lucas sequence with no  primitive divisors. The complete answer to an even more general problem (Lehmer sequences instead of Lucas sequences) was given by Bilu, Hanrot, Voutier  J. Reine Angew. Math.
539, 75-122 (2001)  and Abouzaid  J. Théor. Nombres Bordx.
18, No. 2, 299--313 (2006) .  To settle the conjecture, it suffices to  look up the  tables in these papers .
