$\newcommand{\hom}{\operatorname{Hom}_R}$The statement of Baer's criterion is an $R$-module $E$ is injective iff for all $I \subset R$ an ideal, $\hom(R,E) \twoheadrightarrow \hom(I,E)$. The ``standard" proof of it goes something like this: use Zorn's lemma (one can use the direct product of a linear chain as the upper bound) to get a maximal $N$ with the property $M \hookrightarrow N \hookrightarrow M'$ lifts $M \rightarrow E$. Then if $N \neq M'$ take $m \in M' \setminus N$ and consider $(N:_R m) = I$. Then one can lift $N = N+Im \rightarrow E$ to $N+Im \subsetneq N + Rm$, contradicting maximality of $N$, so $N=M'$ and thus $E$ is injective.
My question is this: why must we have $(N:_R m) \neq 0$? If it is, it seems like this proof doesn't offer us a genuine lift to $N+Rm$, because the map on the $Rm$ part must be $0$.
As @Ben West pointed out in the comments, the proof indeed covers the case $(N:_Rm)=0$.
But even if you dont believe this, you can have a look at this argument:
$(N:_Rm)=0$ means that for any $0 \neq r \in R$, we have $rm \notin N$, i.e. $N \cap Rm = 0$ and thus $N + Rm = N \oplus Rm$. Then you can lift $N \to E$ to $N + Rm \to E$ for trivial reasons.