Each of the $4N$ exterior edges is a side of exactly one face and every other of at most two. So the sum of the sizes of faces is equal to at most $2N^2+4N$. Now, each face has to be at least quadrilateral except corner faces which can be triangles but there are at most $4$ of them taking at most $12$ from $2N^2+4N$. So altogether there can be at most: $$\frac{2N^2+4N-12}{4}$$ non-triangular faces and therefore at most: $$\frac{2N^2+4N-12}{4}+4=\frac{(N+1)^2+1}{2}$$ faces in total. Equality is achieved if for each diagonal we alternate directions. @GerryMyerson