A small block of mass m is placed on a bigger block of mass M that...

Let the amplitude of oscillation be $A$.

The angular frequency of the spring-mass system is given by $\omega = \sqrt{\dfrac{k}{M+m}}$.

The maximum downward acceleration of the system occurs at the highest point of oscillation and its magnitude is a_\max = \omega^2 A = \dfrac{k}{M+m}A.

Consider the forces acting on the small block of mass $m$ at the highest point: the downward gravitational force $mg$ and the upward normal reaction $N$ from the bigger block.

The equation of motion for the small block is:
mg - N = m a_\max

For the block of mass $m$ to not lose contact with the block of mass $M$, the normal reaction must be non-negative ($N \ge 0$).

mg - m a_\max \ge 0
\Rightarrow a_\max \le g

Substituting the value of a_\max:
$\dfrac{k}{M+m}A \le g$
$\Rightarrow A \le \dfrac{(M+m)g}{k}$

Thus, the maximum amplitude with which the two blocks can oscillate together is $\dfrac{(M+m)g}{k}$.