You have correctly identified the maximum velocity and the displacement for that velocity. To get the time, you will have to solve a second-order differential equation.
The system diagram is shown in the picture below. Note that positive acceleration $a$ is in the direction of the positive $x$ axis which is upwards, where $x$ is the displacement of the string. The spring is at rest for $x=0$.
If coordinate system origin was placed at a different position, e.g., at the lowest point of the spring, the differential equation would have an affine component. That is not a big problem, but it is better not to have affine components.
Newton's second law of motion for the above diagram is:
$$ma = k\Delta x - mg$$
where $\Delta x = 0 - x$. Since acceleration is second derivative of distance (displacement), the above equation can also be written as:
$$\frac{d^2}{dt^2} x(t) + \frac{k}{m} x(t) = -g$$
The solution to this second-order differential equation with initial conditions $x(0) = x_0$ and $\dot{x}(0) = \dot{x}_0$ is:
$$x(t) = -\frac{mg}{k} \bigl( 1 - \cos(\omega_0 t) \bigr) + x_0 \cos(\omega_0 t) + \frac{\dot{x}_0}{\omega_0} \sin(\omega_0 t), \qquad t \geq 0$$
where natural frequency is defined as
$$\omega_0 = \sqrt{\frac{k}{m}} \quad \text{[rad/s]}$$
Note that $x_0$ is initial displacement and $\dot{x}_0$ is initial velocity, which are both zero in your example.
The velocity $v$ is defined as first derivative of distance (displacement):
$$v(t) = \frac{d}{dt}x(t) = -\frac{g}{\omega_0} \sin(\omega_0 t) - x_0 \omega_0 \sin(\omega_0 t) + \dot{x}_0 \cos(\omega_0 t), \qquad t \geq 0$$
By taking into account that $x_0 = 0$ and $\dot{x}_0 = 0$ the above two equations are simplified into
$$x(t) = -\frac{mg}{k} \bigl( 1 - \cos(\omega_0 t) \bigr), \qquad v(t) = -\frac{g}{\omega_0} \sin(\omega_0 t)$$
It is obvious now that the absolute maximum velocity
$$v_\text{max} = \frac{g}{\omega_0} \quad \text{[m/s]}$$
occurs when $\omega_0 t = n \frac{\pi}{2}$ where $n$ is any positive odd integer. Therefore, the maximum velocity occurs at $t_\text{max} = n \frac{\pi}{2} \frac{1}{\omega_0}$ for which the spring displacement is $x_\text{max} = -\frac{mg}{k}$.
When you plug the OP's numbers it follows $v_\text{max} = 10\text{ m/s}$, $x_\text{max} = 10\text{ m}$, and $t_\text{max} = \frac{\pi}{2}\text{ s}$.
It should be noted that the mass will permanently oscillate on the spring since there is nothing in the system that will dissipate the energy.
