Solution:

Color the rows of the square black and white alternately, so the top and bottom rows are black. Then each $2 \times 2$ tile covers two cells of each color, and each $3 \times 3$ tile covers six of one color and three of the other. In particular, if only $2 \times 2$ and $3 \times 3$ tiles are used, the difference between the number of black and white cells covered is divisible by $3$. But the entire board has $23$ more black than white cells (if the bottom row were removed, the colors would be equally represented). So there must

be at least one $1 \times 1$ tile.

To construct the required tiling with only one $1 \times 1$ tile, first use $3 \times 3$ tiles to build a $9 \times 12$ rectangle and $2 \times 2$ tiles to build a $2 \times 12$ rectangle. Join these two rectangles to form an $11 \times 12$ rectangle. Then use four copies of this $11 \times 12$ rectangle, together with the $1 \times 1$ square, to build a $23 \times 23$ square as shown in the diagram.