How can i ensure that the sudoku board generates just one unique solution? Does it do it already?

Question

I've been trying to modify TechWithTim's Sudoku solver. I've got it so that it generates a random board, but I am not sure how to ensure that there will be just one unique solution to the puzzle. Any ideas on how I can implement this?

import random


n = 3
m = n**2


def make_board(bo):
    possible_answers = [_ for _ in range(1, m+1)]
    # print(possible_answers)
    find = find_empty(bo)
    if not find:
        return True
    else:
        row, col = find

    random.shuffle(possible_answers)

    for i in possible_answers:
        #  i = random.randint(1, 9)
        if valid(bo, i, (row, col)):
            bo[row][col] = i

            if make_board(bo):
                return True

            bo[row][col] = 0

    return False


counter = 1


def solve(bo):
    global counter
    find = find_empty(bo)
    if not find:
        counter += 1
        return True
    else:
        row, col = find

    for i in range(1, m+1):
        if valid(bo, i, (row, col)):
            bo[row][col] = i

            if solve(bo):
                return True

            bo[row][col] = 0

    return False


def valid(bo, num, pos):
    # Check row
    for i in range(len(bo[0])):
        if bo[pos[0]][i] == num and pos[1] != i:
            return False

    # Check column
    for i in range(len(bo)):
        if bo[i][pos[1]] == num and pos[0] != i:
            return False

    # Checks square
    box_x = pos[1] // n
    box_y = pos[0] // n

    for i in range(box_y * n, box_y * n + n):
        for j in range(box_x * n, box_x * n + n):
            if bo[i][j] == num and (i, j) != pos:
                return False

    return True


def print_board(bo):
    for i in range(len(bo)):
        if i % n == 0 and i != 0:
            print("- - - - - - - - - - - - - ")

        for j in range(len(bo[0])):
            if j % n == 0 and j != 0:
                print(" | ", end="")

            if j == m-1:
                print(bo[i][j])
            else:
                print(str(bo[i][j]) + " ", end="")


def find_empty(bo):
    for i in range(len(bo)):
        for j in range(len(bo[0])):
            if bo[i][j] == 0:
                return i, j  # return row, column

    return None


def count_squares(bo):
    count = 0
    for box in bo:
        for num in box:
            if num != 0:
                count += 1

    return count

This function along with make_board() and solve() are used to generate the board.

def generate_board(diff):
    """generates a board based on difficulty specified"""
    empty_board = [[0 for _ in range(m)] for _ in range(m)]
    global counter

    difficulty_levels = {"Easy": 30, "Medium": 25, "Hard": 20}
    difficulty = difficulty_levels[diff]

    make_board(empty_board)
    counter = 1
    while count_squares(empty_board) > difficulty:
        x = random.randint(0, m-1)
        y = random.randint(0, m-1)
        if valid(empty_board, empty_board[x][y], (x, y)):
            empty_board[x][y] = 0

    return empty_board


board = generate_board("Hard")

print_board(board)
print("\n\n")
solve(board)
print_board(board)