It depends on what you consider a plane drawing to be, but how about something like:
Put a vertex at $(0,0)$ and one at $(1,x)$ for every $x\in\mathbb R$, with a straight edge going between it and $(0,0)$.
Naively this seems to satisfy the requirements of a planar graph drawing: No point lies on two edges (except endpoints), no vertex lies on an edge it is not an endpoint of, every edge is represented by a continuous path between its endpoints.
On the other hand: A problem with this is that if we consider an abstract graph a topological space (with one point per vertex and a copy of the unit interval for each edge, stitched together in the obvious way), then the topology of the above drawing as a subspace of $\mathbb R^2$ is not the same as that of the graph.
Indeed if we define a "plane drawing" of a graph to be a subset of $\mathbb R^2$ which (with the subspace topology) is homeomorphic to the graph, then there can't be any uncountable planar graph, simply because there's no uncountable discrete subset of $\mathbb R^2$ that can be the image of the vertices.
On the other other hand: There are some topological subtleties hidden beneath "stitched together in the obvious way" here. Actually, as soon as there's a node with countable degree the most obvious way of stitching together (resulting in a quotient space) gives something that is not homeomorphic to a straightforward drawing of the graph -- such as a node at $(0,0)$ and one at $(1,n)$ for every $n\in\mathbb N$, with a straight edge to $(0,0)$. This can be fixed, though, by definig the stiching in a slightly more ad-hoc way which gives a different topology on the abstract graph.
It sounds like you're considering an iterative construction which keeps embedding a binary tree inside a larger binary tree, always growing a bit in each direction. If so, this actually does stay countable: picking an arbitrary node and "rotating" everything around that node, what you have is just three copies of the usual (one-way-)infinite binary tree connected at that chosen node.
Of course it's a bit difficult to prove this, given that the definition you have is still rather vague. But I can't see a way of interpreting what you have in mind which doesn't lead to a countable graph. Certainly any time I define a sequence of finite graphs $G_1\subseteq G_2\subseteq G_3\subseteq ...$, their union $$\bigcup_{i\in\mathbb{N}}G_i$$ is countable, so if you want to wind up with something uncountable your construction has to be substantially more complicated.
Best Answer
Here is one explicit way of producing continuum-many pairwise nonisomorphic (simple, undirected, loopless) graphs.
Start with a $3$-cycle among vertices labelled $-2,-1,0$. To the vertex $-1$ attach a tail (= path) of length $1$ and to the vertex $-2$ attach a tail of length $2$. Now, for each positive integer $n$, attach a tail of length $n$ to the vertex $0$, and label the terminal vertex in this tail $n$. (So we have some vertices that we have not labelled, but certainly only countably many of them.)
Now focus on the vertices marked $1,2,3,\ldots$: color the odd numbered ones red and the even numbered ones green. Consider the set of all possible bipartite graphs on this vertex set -- i.e., in which all edges connect red to green. There are clearly continuum-many: for instance even the independent choices of including / not including the edges $1--2$, $1--4$, $1--6$, and so forth gives a set of continuum cardinality. I claim that all of these graphs are pairwise nonisomorphic. Indeed, the only three cycle in any of them is formed by the vertices $-2,-1,0$, and it follows easily from this that any isomorphism between any two of these graphs carries $0$ to $0$, $-1$ to $-1$ and $-2$ to $-2$. From this it follows that for all $n \in \mathbb{Z}^+$, an isomorphism takes the vertex labelled $n$ to the vertex labelled $n$ and thus every vertex is fixed.
(By the way, I don't find the argument you suggest to be at all valid. I strongly suspect there are first order theories such that the number of nonisomorphic models of finite size $n$ grows at least exponentially with $n$ but for which there are only countably many isomorphism classes of countable models.)
Added: I want to make sure that the paths from $0$ to $n$ stay pairwise nonisomorphic when edges are added, so it's best to modify the construction slightly. For instance, call the penultimate vertex in the path $p_n$ and attach one more length one tail at $p_n$, so that now $p_n$ has degree $3$.