It's NP-complete, via a modified version of the reduction Wigderson used to prove that Hamiltonicity of maximal planar graphs is NP-complete.
Careful examination of Wigderson's 1982 NP-completeness proof of hardness for Hamiltonian cycles in maximal planar graphs (http://www.math.ias.edu/avi/node/820) shows that the instances produced by his reduction have the property that there exists an edge $e$ such that either there exists a Hamiltonian cycle through $e$ or there does not exist any Hamiltonian cycle at all. For instance, $e$ can be chosen to be one of the edges in one of Wigderson's $M$-gadgets.
Let $G$ be a hard instance constructed in this way, and embed $G$ so that the edge $e$ belongs to the outer triangle of the embedding. Connect many copies of this embedded graph so that their $e$-edges form a cycle, and make the result maximal planar again by adding two more vertices, one on each side of this cycle, connected to all the exposed vertices of the copies of $G$. Let the number of copies be $c$, and call the resulting graph $H$. Let $n$ be the number of vertices in $G$.
Our hard instance for the largest common subgraph will be the pair $(H,B)$ where $B$ is a bipyramid with the same number of vertices as $H$. Thus, an optimal common subgraph will have to pair all of the vertices. If we make $c$ large enough, the subgraph will necessarily pair the apexes of the bipyramid with the two added vertices in $H$, because their degrees ($c$ and $2c$) will be sufficiently higher than every other vertex in $H$, so that adding these degrees to the solution size will make up for any disruption elsewhere caused by this pairing.
If $G$ is Hamiltonian, then the common subgraph formed by matching the Hamiltonian cycle (minus $e$) in the copies of $G$ to the equator of the bipyramid will have $c(n+2)$ edges, $c(n-1)$ for the equator and $3c$ for the apexes. If $G$ is not Hamiltonian, then (for large enough choices of $c$ that the optimal solution pairs the apexes correctly) any common subgraph will have fewer edges: still $3c$ at the apexes but fewer than $c(n-1)$ elsewhere. So testing whether the common subgraph of $H$ and $B$ has at least $c(n+2)$ edges is NP-complete.