First we show that if we have a counter-example, we can find a counter-example which is connected.
Let $G$ be a planar graph with connected components $G_1$, $G_2$, ..., $G_k$. Select nodes $x_i\in G_i$, and create a new graph $G'$ defined by starting with $G$ and then adding edges $\{x_1,x_2\},\{x_2,x_3\},...,\{x_{k-1},x_k\}$. I'll leave it to you to prove that $G'$ is still planar, and, if $G$ is a counter-example to your theorem, then so is $G'$.
So, you've shown there can be no connected counter-example with $|G|=12$, and hence, by this argument, there can be no counter-example with $|G|=12$.
[Effort to prove via induction elided.]
As Graphth noted above, your argument can be used for $n=|G|\geq 12$, too.
$$m\leq 3n-6$$
$$\sum_{x\in G} deg(x) = 2m \leq 6n-12$$
But if $G$ is a connected counter-example, then $\sum_{x\in G} deg(x) \geq 8(n-5) + 5$
So $$8n-35 \leq 6n-12$$
or
$$2n\leq 23$$
Note that connectedness is more than you really need, you just need to know that $deg(x)\geq 1$ for all nodes $x\in G$. So it's very easy to take any counter-example to a counter-example where no nodes are isolated because adding an edge from an isolated node to another node does not break the planar nature of the graph.