Here is a planar graph for $12$ nodes:
![enter image description here](
The thicker red edges are the onesthat I changed their orientations. Green edges are the ones that are incident to node $12$.
From here, I can say that for $13$ and $14$ nodes, it can be still planar. $13$ is similar to $11$ since $13$th node is just adjacent to $1$ and for $14$, we can add $14$ to outer face in which $2$ and $7$ reside.
For $15$, it is not trivial. But there is a $K_{3,3}$ subgraph using subdivions as the following:
![enter image description here](
Then by Kuratowski's Theorem, it is not planar.
For $16$ nodes, I am sure that it cannot be planar because the nodes $1,2,4,8,16$ will all be connected to each other so we have $K_5$ as a subgraph. Then by Kuratowski's Theorem, it is not planar.