Graph is planar if we can draw the graph on a 2-D plane without overlapping edges.

For $K_{1,n}$ are palanar because we can draw planar graph like,

For $K_{2,n}$ are also planar, we can draw them like

We know that $K_{3,3}$ is non-planar. So for every $m>=3$ and $n>=3$ we get $K_{3,3}$ as sub-graph so it is not planar. [According to Kuratowski's Theorem]

Kuratowski's Theorem states that a graph is planar if and only if it contains no sub-graph of $K_5$ or $K_{3,3}$