Your question has been answered, but I thought I'd add something. For any space $X$, $H_1(X,\mathbb{Z})$ is just the abelianization of $\pi_1(X,*)$. So, to find an example of a space with zero first homology but non-trivial fundamental group, you just need to find a space $X$ with $\pi_1$ equal to its commutator subgroup $[\pi_1,\pi_1]$ (such groups are called perfect groups). Then $H_1(X,\mathbb{Z})\simeq \pi_1^{ab} = \pi_1/[\pi_1,\pi_1] = 0$. These spaces are guaranteed to exist - see, for example, Gabriel's comment.

Your question has been answered, but I thought I'd add something. For any space $X$, $H_1(X,\mathbb{Z})$ is just the abelianization of $\pi_1(X,*)$. So, to find an example of a space with zero first homology but non-trivial fundamental group, you just need to find a space $X$ with $\pi_1$ equal to its commutator subgroup $[\pi_1,\pi_1]$ (such groups are called perfect groups). Then $H_1(X,\mathbb{Z})\simeq \pi_1^{ab} = \pi_1/[\pi_1,\pi_1] = 0$. These spaces are guaranteed to exist - see, for example, Gabriel's comment.

An example of such a space $X$, as has already been mentioned, is the homology 3-sphere. This is the quotient of $S^3$ (viewed as the group of unit quaternions) by the binary icosahedral group $2I$, a perfect (sub)group. In fact, this quotient map is a covering and exhibits $S^3$ as the universal cover of $X$. Hence $\pi_1\simeq 2I$ and $H_1 = 0$.