The case $z=0$ is trivial, just choose $10^k$. We want: \[ n^9 \equiv zzz\dots z = z\frac{10^{k+1}-1}{10-1} \equiv \frac{-z}{9} \mod 10^k\]If $z \in \{ 1,3,7,9 \}$ then $\gcd(z,10)=1$ so we may choose: \[ n = \left(-z\cdot 9^{\varphi(10^k)-1}\right)^{9^{\varphi(\varphi(10^k))-1}} \equiv \left( \frac{-z}{9} \right)^{\frac{1}{9}} \]This is valid since $\gcd(9, \varphi(10^k)) = \gcd(9,\varphi(\varphi(10^k))) = 1$. If $\gcd(z,10) > 1$ then let $p \in \{ 2,5 \}$ divide $z$. We must have: \[ n^9 \equiv z\cdot \overline{111\cdots1}_{k} \mod 10^k \implies p \, \vert \, n^9 \implies \min(p^9,p^k) \, \vert \, z \]This is absurd for $k\geq 4$ since $z$ is too small. Thus the only possible digits are $\{ 0,1,3,7,9 \}$