Deceptively simple mass-spring problem?
You can certainly do it by conservation of energy, but I doubt that will give you the intuitive feel that you are looking for.
So, instead, let's walk through the sequence of event when you let go of the hanging mass ($m_1$).
The hanging mass ($m_1$) starts to fall; the spring begins to extend. It is now exerting a small force up-slope on $m_2$ and a small upward force on $m_1$, but the amount of force is less than either the weight of $m_1$ or the static friction on $m_2$. (Here I assume the pulley is light and low friction so the tensile forces are the same at both ends.) As a result, $m_1$ is picking up speed downward.
As $m_1$ accelerates downward the spring stretches more and the two forces grow. We'll end this "stage" at the point where the force from the spring equals the weight of $m_1$. Keep in mind that $m_1$ has been accelerating downward until the end of this stage. I'll assume the force is still less than the static friction or the problem isn't interesting.
The spring keeps stretching (and the tension keeps growing, so it now exceeds the weight of $m_1$) because $m_1$ is still moving down; however, $m_1$ now has a upward acceleration: it is slowing its downward rush.
If the sliding block doesn't budge, then the maximum tensile force will occur when $m_1$ gets as far down as possible, which means when its speed drops to zero - simply because at that point it turns around and the spring starts getting less extended. This force has a very simple relationship to the weight of $m_1$.
Can you see what that maximum force must be?
Got it.
As $m_1$ drops, gravitational potential energy is converted to spring potential energy:
$$m_1gy=\frac12 k\ell^2$$ Where $y=\ell$ because the string doesn't stretch.
The maximum drop is: $$y_\textrm{max}=\frac{2m_1g}{k}$$ Maximum tension in the spring then is: $$T_\textrm{max}=ky_\textrm{max}=2m_1g$$ For $m_2$ to move: \begin{align}2m_1g &\gt 2m_2g\sin\theta\\ \implies m_1 &\gt m_2\sin\theta\end{align}
If that condition isn't met, $m_2$ won't move and $m_1$ will start moving upwards again and enter an oscillatory motion.
If $m_1$ is too small. The mass fall down and bounces back up. The force on the masses is highest at the lowest point of the bounce. A mass that is just large enough will also bounce back but at the lowest point of the bounce the force will just be large enough to move $m_2$. So we have to find an expression for the force at the lowest point.
Using conservation of energy you can find how far the mass falls by equating the loss in gravitational energy with the gain in energy stored in the spring. From there I guess you should be able to figure it out