Example 2 In space, there is a convex polyhedron, all of whose vertices are lattice points (each vertex has three integer coordinates). In addition, there are no other integer points inside the polyhedron, on its faces, or on its edges. What is the maximum number of vertices this convex polyhedron can have?
Analysis and Solution First, we take the coordinates of the eight grid points as $(i, j, k)(i, j, k \in\{0,1\})$, and with these 8 points as vertices, we can get a unit cube, which obviously meets the requirements of the problem, so the minimum value sought is no less than 8. On the other hand, the three coordinates of each grid point can be divided into the following eight categories according to their parity:
(odd, odd, odd), (odd, odd, even), (odd, even, odd), (odd, even, even),
(even, odd, odd), (even, odd, even), (even, even, odd), (even, even, even).
If this polyhedron has 9 vertices, by the pigeonhole principle, there must be two vertices belonging to the same category of grid points, then the midpoint of the line connecting these two points is also a grid point, and by the convexity of the polyhedron, this grid point belongs to this polyhedron, which contradicts the given information. Therefore, the number of vertices sought is no more than 8.
In summary, the convex polyhedron has at most 8 vertices.