Does trivial cohomology imply trivial homology? Does $\operatorname{Hom}(A,\mathbb Z) = \operatorname{Ext}^1(A, \mathbb Z) = 0$ imply $A = 0$?

Is there a topological space $X$ such that $H^i(X; \mathbb{Z}) = 0$ for all $i > 0$, but $H_n(X; \mathbb{Z}) \neq 0$ for some $n > 0$?

In his answer to the question Is homology determined by cohomology?, Qiaochu Yuan points out that for a Moore space $X = M(A, n)$,

$$\begin{align*} H^n(X; \mathbb{Z}) &\cong \operatorname{Hom}(A, \mathbb{Z})\\ H^{n+1}(X; \mathbb{Z}) &\cong \operatorname{Ext}^1(A, \mathbb{Z}) \end{align*}$$

and $H^i(X; \mathbb{Z}) = 0$ for all $i > 0$. As $\operatorname{Hom}(\mathbb{Q}, \mathbb{Z}) = \operatorname{Hom}(\mathbb{Q}\oplus\mathbb{Q}, \mathbb{Z}) = 0$ and $\operatorname{Ext}^1(\mathbb{Q}, \mathbb{Z}) \cong \mathbb{R} \cong \mathbb{R}\oplus\mathbb{R} \cong \operatorname{Ext}^1(\mathbb{Q}\oplus\mathbb{Q}, \mathbb{Z})$, we see that $M(\mathbb{Q}, n)$ and $M(\mathbb{Q}\oplus\mathbb{Q}, n)$ have the same cohomology, but different homology because $\mathbb{Q} \not\cong \mathbb{Q}\oplus\mathbb{Q}$.

The above shows that, in general, cohomology does not determine homology, but my question is about the specific case where the cohomology is trivial.

Note, if there is an example of a topological space $X$ with trivial cohomology but non-trivial homology, then there is also an example among the Moore spaces. More precisely, if $H_n(X; \mathbb{Z}) \cong A \neq 0$, then $M(A, n)$ also has trivial cohomology but non-trivial homology. Therefore, my initial question is equivalent to the following algebraic question:

Is there a non-trivial abelian group $A$ such that $\operatorname{Hom}(A, \mathbb{Z}) = 0$ and $\operatorname{Ext}^1(A, \mathbb{Z}) = 0$?

If an example of such a group exists, it must not be finitely generated. As such, if there is an $X$ with trivial cohomology but non-trivial homology, it cannot be homotopy equivalent to a compact manifold.

No such groups exist.

Suppose $\operatorname{Ext}(A,\mathbb{Z})=\operatorname{Hom}(A,\mathbb{Z})=0$. First, note that the functor $\operatorname{Ext}(-,\mathbb{Z})$ turns injections into surjections (this follows from the long exact sequence and the fact that $\operatorname{Ext}^2$ always vanishes for abelian groups). So we must have $\operatorname{Ext}(B,\mathbb{Z})=0$ for all subgroups $B\subseteq A$. In particular, this means every finitely generated subgroup of $A$ is free, so $A$ is torsion-free.

Now fix a prime $p$ and consider the short exact sequence $0\to pA\to A\to A/pA\to 0$. Since $A$ is torsion-free, $pA\cong A$. The Ext exact sequence $$\operatorname{Hom}(pA,\mathbb{Z})\to \operatorname{Ext}(A/pA,\mathbb{Z})\to \operatorname{Ext}(A,\mathbb{Z})$$ now tells us that $\operatorname{Ext}(A/pA,\mathbb{Z})=0$. But $A/pA$ is just a direct sum of copies of $\mathbb{Z}/p$, so this implies $A/pA=0$. Thus $pA=A$ for all primes $p$.

This means that $A$ is a divisible torsion-free group, and hence actually a $\mathbb{Q}$-vector space. Since $\operatorname{Ext}(\mathbb{Q},\mathbb{Z})\neq 0$, this implies $A=0$.