Let the square be divided into $n > 1$ polygons. Assume, for contradiction, that none is a triangle. Notice that for one of these polygon, all of its sides are either contained in a side of the square or shared with other polygons. Furthermore, note that it can contain at most one side on each side of the square, and at most one side shared with each polygon. This upper bounds the number of sides of each of the polygons by $(n-1) + 4 = n+3.$ Since the number of sides of the polygons are all distinct and lower-bounded by $4$, we've that their number of sides are $4, 5, 6, \cdots, n+3$ in some order. So the polygon with $n+3$ sides shares one side with all four sides of the square, and also with all of the other polygons. Now, let's observe that polygon with $n+2$ sides. It must lie in one of the four regions "cut off" by the polygon of $n+3$ sides, and therefore can only touch two of the sides of the square. Hence, it has at most $2 + (n-1) = n+1$ sides, absurd. $\square$