Please, read my comment to the question...
My best guess is:
DECLARE @SEARCH_STRING VARCHAR(255) = 'Whatever [!] the string is... It is for...'
DECLARE @counter INT = 0
DECLARE @stringlenght INT = LEN(@SEARCH_STRING)
WHILE (@counter<=@stringlenght)
BEGIN
IF (SUBSTRING(@SEARCH_STRING, @counter,1)='[')
SELECT CONCAT('A ''', SUBSTRING(@SEARCH_STRING, @counter, 1), ''' has been found on ', CONVERT(VARCHAR(3), @counter), '. position.')
SET @counter = @counter + 1
END
Result:
A '[' has been found on 10. position.