Is there a way to get the positional parameters of the script from inside a function in bash?

No, not directly, since the function parameters mask them. But in Bash or ksh, you could just assign the script's arguments to a separate array, and use that.

#!/bin/bash
ARGV=("$@")
foo() {
     echo "number of args: ${#ARGV[@]}"
     echo "second arg: ${ARGV[1]}"
}
foo x y z 

Note that the numbering for the array starts at zero, so $1 goes to ${ARGV[0]} etc.

Another way to get the param of the script with bash is to use the Shell Variables BASH_ARGC and BASH_ARGV

#!/bin/bash
shopt -s extdebug
test(){
  echo 'number of element in the current bash execution call stack = '"${#BASH_ARGC[*]}"
  echo 'the script come with '"${BASH_ARGC[$((${#BASH_ARGC[*]}-1))]}"' parameter(s)'
  echo 'the first is '"${BASH_ARGV[$((${#BASH_ARGV[*]}-1))]}"
  echo 'there is 2 way to get the parameters of this function'
  echo 'the first is to get $1,...,$n'
  echo '$1 = '"$1"
  echo '$2 = '"$2"
  echo 'the second with the use of BASH_ARGC and BASH_ARGV'
  echo 'this function '"${FUNCNAME[0]}"' come with '"${BASH_ARGC[0]}"' parameter(s)'
  echo 'the second is '"${BASH_ARGV[0]}"
  echo 'the first is '"${BASH_ARGV[1]}"
}
essai(){
  test paramtest1 "$3"
}
essai paramessai1 paramessai2 paramessai3