Let $\lambda, \mu$ be two real valued measures in the measurable space $(X, \mathcal{A})$. Bartle's "Elements of Integration" Lemma 8.8 says that, if both are finite measures, then $\lambda \ll \mu$ if, and only if, for every $\varepsilon > 0$, there exists $\delta > 0$ such that, for all $E \in \mathcal{A}$, if $\mu(E)< \delta$, then $\lambda(E)< \epsilon$.
The $(\Leftarrow)$ side of the proof is obvious, because if $\mu(E) = 0$, then $\mu(E)< \delta$ for each $\delta > 0$, which implies $\lambda(E) < \varepsilon$ for each $\varepsilon > 0$. Then we have $\lambda(E) = 0$. This proves that $\lambda \ll \mu$.
The other side of the proof goes as follows: suppose the conclusion of $(\Rightarrow)$ is false. Then there is $\varepsilon > 0 $ such that for each $n \in \mathbb{N}$, there is $E_n \in \mathcal{A}$ with $\mu(E_n)< 2^{-n}$ and $\lambda(E_n) \geq \varepsilon$. Define $F_n = \bigcup_{k=n}^\infty E_n$. Then $F_n$ is a decreasing sequence of sets, $\mu(F_n) \leq 2^{-n + 1}$, $\mu(F_1) \leq 1 < \infty$ and $\lambda(F_1)< \infty$, because $\lambda$ is finite. Then $\mu(\bigcap F_n) = \lim \mu(F_n) = 0$ and $\lambda(\bigcap F_n) = \lim \lambda(F_n) \geq \varepsilon$.
It seems to me that the hypothesis that $\mu$ is finite can be dropped. Is this the case?
Note that after you prove Radon-Nikodym theorem the equivalence is still true. That is, if $\lambda$ is finite and $\mu$ is $\sigma$-finite with $\lambda\ll \mu$, then there is an measurable function $g$ s.t $$\lambda(A)=\int_Ag\ d\mu$$ for all measurable $A$. Now, by finiteness of $\lambda$ it follows that $g$ is integrable. So, by the continuity of the integral you have the direction ($\implies$).
As it seems by your observation the equivalence is still true in the special case where $\lambda$ is finite and $\mu$ is an arbitraty positive measure whereas Radon-Nikodym is not since you can take as $\lambda$ the Lebesgue measure on $(0,1)$ and $\mu$ the counting measure on $(0,1)$.