Assume $A$ is a commutative ring and $M$ an $A$-module such that $\mathrm{Ext}^1_A(N,M)=0$ for all cyclic modules $N$. I wish to prove $M$ is an injective module, so that the above equality holds for all $N$ and not just cyclic modules.
My initial idea was to assume some $N$ contradicts the claim, and take the poset $\{ L \subset N \mid \mathrm{Ext}_A^1(L,M)=0\}$. If we prove that ascending chains in this poset have some maximal element, then Zorn's Lemma implies this poset has a maximal element $L$. If $a\in N\setminus L$, then the short exact sequence $$0\to L \to L+Aa \to (L+Aa)/L \to 0$$ gives, since $(L+Aa)/L$ is cyclic and so $\mathrm{Ext}^1_A((L+Aa)/L,M)=0=Ext^1_A(L,M)$, an associated long exact sequence which proves $L+Aa$ is also in the poset, a contradiction to maximality of $L$.
Unfortunately I'm having trouble proving that if $L_1\subset L_2\subset \cdots$ are in the poset, then $L:= \bigcup_{i=1}^\infty L_i$ is also in the poset. Any help would be greatly appreciated!
Let $A=\mathbb{Z}$. For every rational number $q$ consider the $\mathbb{Z}$-submodule $ \langle q \rangle$ of $\mathbb{Q}$. Then $\bigcup_{q \in \mathbb{Q}} \langle q \rangle = \mathbb{Q}$ and $\mathrm{Ext}_{\mathbb{Z}}^1(\langle q \rangle, \mathbb{Z}) =0$ since $\langle q \rangle$ is a projective $\mathbb{Z}$-module. But $\mathrm{Ext}_{\mathbb{Z}}^1(\mathbb{Q},\mathbb{Z}) \neq {0}$. So the assertion you are trying to prove is false.
To prove your claim you have to check the exactness of $\mathrm{Hom}_A(-,M)$. It follows from the Baer criterion that it suffices to check the exactness of $\mathrm{Hom}_A(-,M)$ on exact sequences of the form $0 \to I \to A \to A/I \to 0$, where $I\subset A$ is an ideal. Now in the long exact $\mathrm{Ext}$ sequence $$0 \to \mathrm{Hom}_A(A/I,M) \to \mathrm{Hom}_A(A,M) \to \mathrm{Hom}_A(I,M) \to \mathrm{Ext}_A^1(A/I, M)\to\cdots$$ $\mathrm{Ext}_A^1(A/I, M)$ vanishes because $A/I$ is a cyclic module.