How can I prove that if $G$ doesn't contain a $K_{3,3}$ or $K_{5}$ subgraph, then $G$ is planar? Or is this statement false?
I'm aware of Kuratowski's Theorem, which states that $G$ is nonplanar if and only if $G$ contains a subgraph that is a subdivision of either $K_{3,3}$ or $K_{5}$. I'm not sure if I can go from here to the result that I'd like to prove. I would greatly appreciate any help.
Your conjecture is false.
Consider the graph $G$ that is $K_5$, but each edge has had a vertex added in the middle. [So it has $5$ vertices of degree $4$ and $10$ vertices of degree $2$. It is of course a subdivision of $K_5$.] This graph is not planar---if we had a planar drawing then just erasing the vertices of degree $2$ would give us a planar drawing of $K_5$ (which is impossible). On the other hand $G$ contains no $K_5$ or $K_{3,3}$ subgraph.