Well, any finitely generated group G of exponent 3 is finite by a classical theorem of Burnside. And since the order of every element is 3, the order of G must be a power of 3 by Cauchy's theorem. It follows that G is a finite nilpotent group. A similar argument shows that the same is true for any finitely generated group of exponent 4. This is unknown for 5, and false for 6. On the other hand, if G is not finitely generated then it probably does not need to be nilpotent (I do not think that the exponent 3 yields a bound on the nilpotency degree, and so the direct product of all finitely generated nilpotent groups of exponent 3 will still have exponent 3, but will not be nilpotent).

Well, any finitely generated group G of exponent 3 is finite by a classical theorem of Burnside. And since the order of every element is 3, the order of G must be a power of 3 by Cauchy's theorem. It follows that G is a finite nilpotent group. A similar argument shows that the same is true for any finitely generated group of exponent 4. This is unknown for 5, and false for 6.

Correction: it seems (see the answer of Primoz above) that any group of exponent 3 is nilpotent, altough it can be infinite if it not finitely generated.