Two blocks having masses m_1 and m_2 are connected by a spring with...

Let $v$ be the common velocity of the two blocks at the instant of maximum elongation.

By conservation of linear momentum:
$m_2 v_0 = (m_1 + m_2)v \Rightarrow v = \dfrac{m_2 v_0}{m_1 + m_2}$

By conservation of mechanical energy:
\dfrac{1}{2} m_2 v_0^2 = \dfrac{1}{2} (m_1 + m_2)v^2 + \dfrac{1}{2} k x_\max^2

Substituting the value of $v$:
\dfrac{1}{2} m_2 v_0^2 = \dfrac{1}{2} (m_1 + m_2) \left( \dfrac{m_2 v_0}{m_1 + m_2} \right)^2 + \dfrac{1}{2} k x_\max^2

m_2 v_0^2 = \dfrac{m_2^2 v_0^2}{m_1 + m_2} + k x_\max^2

k x_\max^2 = m_2 v_0^2 \left( 1 -\dfrac{m_2}{m_1 + m_2} \right)

k x_\max^2 =\dfrac{m_1 m_2}{m_1 + m_2}$v_0^2$

x_\max = v_0 \left[\dfrac{m_1 m_2}{(m_1 + m_2)k} \right]^{1/2}