By considering centers of the hexagons, we get the following equivalent problem:
"Two vertices $\alpha$ and $\beta$ of the triangular lattice are called equivalent if $\alpha - \beta$ is equal to sum of finitely many vectors from the set:
$$A = \{\pm(n\vec{i} - m\vec{j}), \pm(n\vec{k} + m\vec{i}), \pm(n\vec{j} + m\vec{k})\}$$
Where $\vec{i} = (1,0)$, $\vec{j} = (-\frac{1}{2}, \frac{\sqrt{3}}{2})$ and $\vec{k} = (\frac{1}{2}, \frac{\sqrt{3}}{2})$ according to the normal Cartesian coordinates.

What is the maximum number of nonequivalent vertices?
Note that $\vec{k} = \vec{i} + \vec{j}$, thus we can determine vectors of set $A$ using only $\vec{i}$ and $\vec{j}$:
$$A = \{\pm(n\vec{i} - m\vec{j}), \pm((n+m)\vec{i} + n\vec{j}), \pm(m\vec{i} + (m+n)\vec{j})\}.$$
Since the first vector $n\vec{i} - m\vec{j}$ is the difference of the two others, two vertices $\alpha$ and $\beta$ are equivalent iff their difference $\alpha - \beta$ can be written as a linear combination of vectors $(n+m)\vec{i} + n\vec{j}$ and $m\vec{i} + (m+n)\vec{j}$ with integer coefficients.
Note that for integers $a, b, x$ and $y$ the space generated by the vectors $\{a\vec{i} + b\vec{j}, x\vec{i} + y\vec{j}\}$ (using integer coefficients) equals the space generated by vectors $\{(a-x)\vec{i} + (b-y)\vec{j}, x\vec{i} + y\vec{j}\}$. Therefore, we can change former vectors to get simpler vectors to work with. Note that in this change the value of $ay - bx$ (determinant of the matrix formed by vectors $(a, b)$ and $(x, y)$ as rows) is invariant, because $ay - bx = (a-x)y - (b-y)x$. Now, in a similar way to Euclidean Algorithm, if $a \ge x$ we change $\{a\vec{i} + b\vec{j}, x\vec{i} + y\vec{j}\}$ by $\{(a-x)\vec{i} + (b-y)\vec{j}, x\vec{i} + y\vec{j}\}$. Repeating this process several times leads to vectors of the form $s\vec{i} + r\vec{j}$ and $0\vec{i} + t\vec{j}$. It means that we can suppose the coefficient of $\vec{i}$ for one of them is zero. hence, if we want to go from one vertex to another, the difference of the coefficients of $\vec{i}$ must be a multiple of $s$, since $t\vec{j}$ does not affect the coefficient of $\vec{i}$. As a result, for two equivalent vertices, we can first use vector $s\vec{i} + r\vec{j}$ several times to match their first coordinate. Then, we must use $t\vec{j}$ to match the second coordinate. Hence we have $st$ nonequivalent vertices. But
$$st = \det \begin{bmatrix} s & r \\ 0 & t \end{bmatrix}$$
which is invariant during the process.
$$\det \begin{bmatrix} s & r \\ 0 & t \end{bmatrix} = \det \begin{bmatrix} m+n & m \\ n & m+n \end{bmatrix} = m^2 + mn + n^2$$
So the maximum number of nonequivalent vertices is $m^2 + mn + n^2$.