continued fraction calculator and plug in "3" in the square root box. You will then see a list of the convergents for the continued fraction expansion of , and you will notice that every fourth item in the list has an even numerator:
Let be the n'th term in that sequence. Then and the sequence grows by the inductive rules , . You should then be able to show by induction that . That gives an infinite family of integer solutions to the equation . Going back to the original problem, , so that gives infinitely many integers x such that is a square.