In the "The NP-Completeness Column: An Ongoing Guide" number 14, Johnson writes "... Vergis [49]. Transformation from STEINER TREE IN GRAPHS [ND12]...". I don't have access to the Vergis' paper, however a possible reduction can be the following.
Steiner Tree (ST) in graphs problem
Instance: an undirected graph $G = (V,E)$, a subset of the vertices $R \subseteq V$, called terminal nodes; a non-negative integer $k$
Question: is there a subtree of $G$ that includes all the vertices of $R$ (a spanning tree for $R$) and that contains at most $k$ edges?
The problem remains NPC even for planar graphs (M. Garey and D. Johnson. The rectilinear Steiner tree problem is NP-complete).
Given an instance of planar ST, for now suppose that you can assign arbitrary weights to nodes. If $|R|=p$ and $|E|=q$ , you can assign weight $q+1$ to the nodes of $R$ and you can add a middle node to every edge of $E$ and assign weight $-1$ to it. Assign weight $0$ to the remaining nodes in $V \setminus R$. The original graph $G$ has a spanning tree for $R$ with at most $k$ edges if and only if in the transformed graph you can find a subgraph of target weight greater or equal to $W = p(q+1) - k$.
Informally to reach the $p(q+1)$ quantity you must include all the nodes of $R$ in the subgraph, and you must include at most $k$ middle nodes (corresponding to the edges of $G$) that have negative weight -1 to keep them connected.
You can reduce the the max degree of the whole graph to three in this way: if $D = \max\{deg(u_i) \mid u_i \in V\}$ just transform each node $u_i$ to a circular chain of $D$ nodes (and adjust the value of $p$ accordingly). Connect the inbound edges to distinct nodes of the chain (that will have degree 3).
And if you want to use only weights $+1,-1$ then you must:
(A) assign $+1$ to all the nodes of the circular chains corresponding to nodes in $V\setminus R$,
(B) transform every middle node to a linear chain of nodes of weight $-1$ and length $l_E = |V\setminus R|+1$, and
(C) further expand the circular chains (with weight +1) corresponding to the nodes of $R$ to at least length $l_R = l_E( |E| + 1)$; and set target weight $W =pl_R - kl_E$.
Informally, the expanded chains and the new target make the $+1$ weights of the nodes corresponding to $V \setminus R$ (point A) irrelevant.
