Racket, 125125 122 bytes
(for([x(range 1 101)])(define(m n)(=(modulo x n)0))(displayln(cond[(and(m 3)(m 5))"FizzBuzz"]['FizzBuzz][(m 3)"Fizz"]['Fizz][(m 5)"Buzz"][x]'Buzz][x])))
Simplest approach, took some work to get it lower than 130 bytes. Inspired by the Java example.
Pretty-printed code
(for ([x (range 1 101)])
(define (m n)
(= (modulo x n) 0))
(displayln (cond
[(and (m 3) (m 5)) "FizzBuzz"]'FizzBuzz]
[(m 3) "Fizz"]'Fizz]
[(m 5) "Buzz"]'Buzz]
[x])))