Scraping names from multipages without clicking on the next button
The code is pretty much straightforward and understandable, but I would still work on the following stylistic and readability issues:


extract constants. It might be a good idea to extract the URL template and maximum page number into separate constants, or as arguments of a function
put your main execution logic into the if __name__ == '__main__' to avoid it being executed if the module is imported
avoid using bare exception clauses. Be specific about the exceptions you are handling. In this case, NoSuchElementException is a good exception to handle if an item name is not found
use try/finally to safely close the driver if it fails before quitting - this way you would eliminate a situation when you have opened "ghost" browser windows left from failed scraping tries


All things applied:

from selenium import webdriver
from selenium.common.exceptions import NoSuchElementException
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC


def scrape(url, max_page_number):
    driver = webdriver.Chrome()
    wait = WebDriverWait(driver, 10)

    try:
        for page_number in range(1, max_page_number + 1):
            driver.get(url.format(page_number))

            for item in wait.until(EC.presence_of_all_elements_located((By.CSS_SELECTOR, ".info"))):
                try:
                    name = item.find_element_by_css_selector('a.business-name span[itemprop=name]').text
                except NoSuchElementException:
                    name = ''

                print(name)
    finally:
        driver.quit()


if __name__ == '__main__':
    url_template = 'https://www.yellowpages.com/search?search_terms=pizza&amp;geo_location_terms=San%20Francisco%2C%20CA&amp;page={0}'
    max_page_number = 2

    scrape(url_template, max_page_number)