Proof/Counterexample: If $z$ is a complex number and $z\notin \mathbb Q$, then $\mathbb Q(z)=\mathbb Q(z^3,z^5)$.
First, $\mathbb Q(z)\subseteq \mathbb Q(z^3,z^5)$ would be trivial, right? Then we are left with showing $\mathbb Q(z)\supseteq \mathbb Q(z^3,z^5)$.
If $z\notin \mathbb Q$, then for $z=a+bi$ where $a,b\in \mathbb Q$, $b\ne 0$.
$a+bi=(a+bi)^3$
$a+bi=a^3+3a^2bi-3b^2a-b^3i$
$a+bi=(a^3-3b^2a)+(3a^2b-b^3)i$
Then because $a,b\in \mathbb Q$, $(a^3-3b^2a)+(3a^2b-b^3)i$ is contained in $\mathbb Q(z)$. Thus, $\mathbb Q(z)\supseteq \mathbb Q(z^3,z^5)$?
Does it work like this, or am I way off base with this attempt?
No, you are way off base. Your reasoning doesn't make much sense (see my comments above).
In fact you start from the wrong foot. The trivial inclusion is $\mathbb{Q}(z^3,z^5) \subseteq \mathbb{Q}(z)$. Indeed, it's clear that both $z^3$ and $z^5$ can be written as rational fractions of $z$: they're both polynomial in $z$.
To prove the other inclusion $\mathbb{Q}(z) \subseteq \mathbb{Q}(z^3,z^5)$, you need to prove that $z$ can be written in terms of $z^3$ and $z^5$ using only addition, multiplication, division, and multiplication by a rational number. You can do that without using the representation $z=a+bi$, it's purely formal: since $5 \times 2 - 3 \times 3 = 1$, you find that: $$z = z^1 = z^{5 \times 2 - 3 \times 3} = \frac{(z^5)^2}{(z^3)^3}.$$