Update: This proof no longer works as is for the new question about positive integers -- I suspect it could be fixed, but I won't bother since the other one is simpler and not affected by the change.
To generate $1$, we need to have $z^k=a+\mathrm{i}b$, and hence $(\lvert z \rvert^2)^k=a^2+b^2$, with $a=1$ or $b=1$. Since $\lvert z \rvert^2$ is an integer, $k$ cannot be even, for otherwise we would have two squares that differ by $1$, which is only possible if $\lvert z \rvert=1$, which is obviously not a solution.
So $k$ must be odd. Now consider $(x + \mathrm{i}y)^k$ for $k$ odd (and still $a=1$ or $b=1$). Then all terms in the real part contain a factor $x$ and all terms in the imaginary part contain a factor $y$. Since one of these parts is $1$, it follows that one of $x$ and $y$ is $\pm 1$. They cannot both be $\pm 1$, since that is obviously not a solution. Denote the one that isn't $\pm1$ by $w$. (Also clearly $w\neq0$.)
Now consider $(x + \mathrm{i}y)^k$ for arbitrary $k$. All terms in the real and imaginary parts are either $\pm 1$ or multiples of $w$. Thus the only residues modulo $w$ that occur are $\pm1$ and $0$. It follows that $\lvert w \rvert \le 3$, i.e. $w=\pm3$ or $w=\pm2$. Both cases are easily excluded: With $w=\pm3$, both parts of $z^2$ are even, and hence we cannot generate any odd integers other than $\pm1$ and $\pm3$. With $w=\pm2$, the residues modulo 5 of the real and imaginary parts are periodic in $k$ with period $4$ (period $2$ if we drop their sign) and none of the residues in the period is $0$, and hence we cannot generate any multiples of $5$. $\Box$
P.S.: I just realized that I killed this proof when I proposed to save the question by including $z^0$ :-). Fortunately the proof can be saved just as easily; just replace $1$ by $-1$.
P.P.S.: This avenue has now also been closed by the restriction to positive integers.