$m$ and $n$ are two nonnegative integers. In the Philosopher's Chess, The chessboard is an infinite grid of identical regular hexagons and a new piece named the Donkey moves on it as follows:

Starting from one of the hexagons, the Donkey moves $m$ cells in one of the $6$ directions, then it turns $60$ degrees clockwise and after that moves $n$ cells in this new direction until it reaches it's final cell.

At most how many cells are in the Philosopher's chessboard such that one cannot go from anyone of them to the other with a finite number of movements of the Donkey?

[i]Proposed by Shayan Dashmiz[/i]
1. **Coordinate System Setup**:
   Set up a coordinate system in which the \( y \)-axis makes a \( 60^\circ \) angle with the \( x \)-axis. Let \( g = \gcd(m, n) \) so that \( m = gm' \) and \( n = gn' \) where \( \gcd(m', n') = 1 \).

2. **Possible Moves**:
   From a hexagon at \((x, y)\), the Donkey can move to one of six possible positions \((x', y')\):
   \[
   (x \pm (m + n), y \mp n), \quad (x \pm n, y \pm m), \quad (x \pm m, y \mp (m + n))
   \]
   Notice that \( g \mid (x' - x) \) and \( g \mid (y' - y) \).

3. **Congruence Condition**:
   Also, we have:
   \[
   m' \frac{(x' - x)}{g} - n' \frac{(y' - y)}{g} \equiv 0 \pmod{m'^2 + m'n' + n'^2}
   \]
   regardless of the choice of move.

4. **Sufficiency of the Condition**:
   This condition is sufficient to guarantee that we can move from \((x, y)\) to any \((x', y')\). If we make \( a, b, c \) moves of each type, respectively, we want to solve:
   \[
   \begin{cases}
   x' - x = a(m + n) + bn + cm \\
   y' - y = -an + bm - c(m + n)
   \end{cases}
   \]
   for \( a, b, c \) in the integers.

5. **Solving the System**:
   We can arbitrarily set \( c = 0 \) and after dividing through by \( g \), note that by Cramer's Rule and the fact that:
   \[
   (m' + n')(m) - n'(-n') = m'^2 + m'n' + n'^2 \mid m' \frac{(x' - x)}{g} - n' \frac{(y' - y)}{g}
   \]
   the system has an integer solution.

6. **Maximum Number of Cells**:
   The maximum number of cells is \( m^2 + mn + n^2 \). If we had any more, by the Pigeonhole Principle, some pair \((x_1, y_1), (x_2, y_2)\) would satisfy \( g \mid (x_1 - x_2) \), \( g \mid (y_1 - y_2) \), and:
   \[
   m'^2 + m'n' + n'^2 \mid m' \frac{(x_1 - x_2)}{g} - n' \frac{(y_1 - y_2)}{g}
   \]
   so we could move from one to the other.

7. **Attainability**:
   This is attainable by choosing a set of points where for each pair \((p, q)\) of proper residues modulo \( g \), we include \( m'^2 + m'n' + n'^2 \) points \((x_i, y_i)\) where \( x_i \equiv p \pmod{g} \), \( y_i \equiv q \pmod{g} \), and:
   \[
   m' \frac{(x_i - p)}{g} - n' \frac{(y_i - q)}{g}
   \]
   takes on all \( m'^2 + m'n' + n'^2 \) possible values modulo \( m'^2 + m'n' + n'^2 \) (which is possible by Bézout's identity and the fact that \( \gcd(m', n') = 1 \)).

8. **Geometric Interpretation**:
   Consider the infinite equilateral lattice of points reachable from a single starting point. We want the "area" of the parallelogram formed by the vectors \((n, m)\) and \((m + n, -n)\). This area is:
   \[
   \left| \begin{array}{cc}
   n & m \\
   m + n & -n
   \end{array} \right| = m^2 + mn + n^2
   \]

The final answer is \( \boxed{ m^2 + mn + n^2 } \)