Every simple planar graph with $\delta\geq 3$ has an adjacent pair with $\deg(u)+\deg(v)\leq 13$ 
Claim: Every simple planar graph with minimum degree at least three has an edge $uv$ such that $\deg(u) + \deg(v)\leq 13$.  Furthermore, there exists an example showing that $13$ cannot be replaced by $12$.

This seems related to Hard planar graph problem.
By Euler's formula, we know that $|V|-|E|+|F|=2$.  Rearranging, we have $|F| = 2+|E|-|V|$
As $\delta\geq3$ we know $|V|\geq 4$ and $|E|\geq 6$ and since each face has at least 3 edges bounding it and each edge is on the boundary of at most 2 faces we have then that $3|F|\leq 2|E|$.  Substituting into Euler's formula, we have $$2|E|\geq 3|F| = 3(2+|E|-|V|)\Rightarrow 3|V|-6\geq |E|$$
From this we have from the handshaking lemma that avgdegree = $\sum\frac{\deg(v)}{|V|} = 2\frac{|E|}{|V|} \leq 6 - \frac{12}{|V|}\lneq 6$.  And so, the average degree of a simple planar graph is strictly less than 6.
Suppose there was a smallest counterexample, I.e., some graph such that every edge $uv$ has $\deg(u)+\deg(v)>13\dots$

It is at this point that I feel as though I am stuck.  It seems that you should be able to reach a contradiction about either the average degree, the planarity, or $\delta\geq 3$.  In the hint in the linked similar problem, they suggest to show that you can find a stricter bound for the minimum degree of the smallest counterexample from below, but it doesn't seem to apply here.  They then suggest to show that a substantial proportion of the vertices are of low degree (in this problem would translate to degree less than 6), and then for the punchline to show that the number of edges is enough that there are more than #of vertices of high degree choose 2, meaning there must necessarily be an edge between two vertices of low degree thus proving the claim.

This may be better as a simple comment, but I lack the reputation.
It can be assumed that we are working with a triangulation, but we need to be careful about which edges we add when triangulating. Suppose that the graph has minimum degree three and there is no edge $uv$ with $\deg(u) + \deg(v) \leq 13$. Now suppose there were a face of degree at least four. Let $x$ be the vertex of minimum degree on that face. In any case, both neighbours of $x$ on the face will have degree at least $7$. Add an edge between those two vertices and continue this process until you are left with only triangular faces.
I stole this trick from https://faculty.math.illinois.edu/~west/pubs/dischnew.pdf where it is used at the start of Lemma 3.2. Incidentally, that lemma proves something stronger than what is required here: Every plane graph with minimum degree $3$ has an edge with weight at most $11$ or a $4$-cycle through two degree $3$ vertices and a common neighbour of degree at most $10$.
