Or to put it differently: The work-energy theorem holds for any force, i.e., from the equation of motion
$$m \dot{\vec{v}}=\vec{F},$$
where ##\vec{F}## can be an arbitrary function of ##\vec{x}##, ##t##, ##\dot{\vec{x}}##(, and even higher time derivatives of ##\vec{x}##, but this usually leads to other trouble). Now multiply this with ##\vec{v}## (using the scalar product). Then you get
$$m \vec{v} \cdot \dot{\vec{v}}=\vec{v} \cdot \vec{F}.$$
Now the left-hand side is a total time derivative, i.e.,
$$m \vec{v} \cdot \dot{\vec{v}}=\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{m}{2} \vec{v}^2 \right)=\dot{E}_{\text{kin}}=\vec{v} \cdot \vec{F}.$$
Now integrate this wrt. time over a time interval ##(t_1,t_2)## this gives the work-energy theorem,
$$E_{\text{kin}}(t_2)-E_{\text{kin}}(t_1)=\int_{t_1}^{t_2} \mathrm{d} t \vec{v} \cdot \vec{F}=W(t_1,t_2),$$
i.e., the change in kinetic energy is given by the work done in the said time interval.
This doesn't help much, if you don't know the solution of the equation of motion, and it's thus a pretty dull result. This drastically changes, if ##\vec{F}## is a potential field, i.e., if it's a function of ##\vec{x}## only, and it can be written as a gradient of a scalar field ##V##. Usually one defines
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
Using this in the work-energy theorem leads to
$$W(t_1,t_2)=-\int_{t_1}^{t_2} \mathrm{d} t \dot{\vec{x}} \cdot \vec{\nabla} V(\vec{x}) = -\int_{t_1}^{t_2} \mathrm{d} t \frac{\mathrm{d}}{\mathrm{d} t} V(\vec{x})=-V(\vec{x}_2)+V(\vec{x}_1),$$
where ##\vec{x}_2=\vec{x}(t_2)## and ##\vec{x}_1=\vec{x}(t_1)##. So you have from the work energy theorem
$$E_1=E_{\text{kin} 1}+V(\vec{x}_1)=E_2=E_{\text{kin} 2} +V(\vec{x}_2),$$
i.e., the then definable total energy,
$$E=E_{\text{kin}}+V(\vec{x})=\text{const},$$
is conserved for any solution of the equation of motion, i.e., you don't need to know the specific equation of motion, to have some information from enery conservation.
For a force in 1D motion, ##F(x)=-k x## you have ##V(x)=k x^2/2## and thus
$$\frac{m}{2} v^2 + \frac{k}{2} x^2=\text{const}.$$
I'm not sure about the initial conditions in the OP. I guess, at ##t=0## you expanded the spring by ##x_0>0## from the equilibrium position ##x=0## and then realease it, i.e., your initial conditions are ##x(0)=x_0## and ##v(0)=0##. Then the energy-conservation law tells you that
$$E=\frac{m}{2} v^2 + \frac{k}{2} x^2=E_0=\frac{k}{2} x_0^2.$$
This tells you, without solving for the equations of motion that you always must have ##|x|<x_0## and that the maximal velocity is for ##x=0##, and that this maximal speed is given by
$$\frac{m}{2} v_{\text{max}}^2=E_0=\frac{k}{2} x_0^2 \; \Rightarrow \; |v_{\text{max}}|=\sqrt{\frac{k}{m}} x_0.$$
For ##|x|=x_0## you always have ##v=0##. This tells you that the mass oscillates between the points ##x=x_0## and ##x=-x_0##, and at this turning points ##\dot{x}=v=0##. So you get a pretty good qualitative picture about the motion, without solving for the equation of motion, just from the energy-conservation law, which holds for the forces, which have a potential in the above described sense.
