Let us restate the problem for ease of reference. The equations are
$\dot{x}(t)=y(t)$
$\dot{y}(t)= - 4 x(t) + y(t)^2$
Physically they describe an anharmonic oscillator with spatial coordinate $x(t)$ and velocity $y(t)$ in the potential
$U = 2 x^2 - \frac{1}{3} x^3$
and the energy
$E = \frac{1}{2} y^2 + U(x)$
The problem is then: given two solutions
$s_{1}(t) = (x_{1}(t), y_{1}(t))$
$s_{2}(t) = (x_{2}(t), y_{2}(t))$
corresponding to the initial conditions
$s_{1}(0) = (x_{10}, y_{10})$
$s_{2}(0) = (x_{20}, y_{20})$
Assuming that $y_{10}$ and $y_{20}$ do not vanish show that the quantity
$d(t) = y_{1}(t) + y_{2}(t)$
becomes $\neq 0$ for times $t>0$ even if it is $= 0$ for $t = 0$, i.e.
$y_{10} + y_{20} = 0$.
We prove it indirectly, assuming $d(t) = 0$ for all times.
The idea is to expand the solution $y(t)$ into a power series in $t$.
$y(t) = y(0) + t \dot{y}(0) + t^2 \frac{1}{2} \ddot{y}(0) + t^3 \frac{1}{6} \frac{\partial ^3y(0)}{\partial t^3} + ...$
In order to have $d(t) = 0$, all coefficients must vanish. These can be expressed through the initial values, and the first coefficients are
$c(0) = (y_{10} + y_{20}) $
$c(1) = - 4 (x_{10} + x_{20}) + (x_{10}^2 + x_{20}^2) $
$c(2) = - 4 (y_{10} + y_{20}) + 2 ( x_{10} y_{10} + x_{20} y_{20} ) $
$c(3) = 2 (y_{10}^2 + y_{20}^2) + 2 ( x_{10}^3+x_{20}^3 ) -12 (x_{10}^2 + x_{20}^2) + 16 (x_{10} + x_{20})) $
Since we have $y_{10} + y_{20} = 0$, $d(t) = 0$ requires
$c(1) = 0 = - 4 (x_{10} + x_{20}) + (x_{10}^2 + x_{20}^2) $
$c(2) = 0 = 2 y_{10} (x_{10} - x_{20}) $
$c(3) = 0 = 4 y_{10}^2 + 2 ( x_{10}^3 + x_{20}^3 ) -12 (x_{10}^2 + x_{20}^2) + 16 (x_{10} + x_{20})) $
From $c(2) = 0$ and $y_{10} \neq 0$ we find $x_{20} = x_{10}$. Hence
$c(1) = 0 = - 4 x_{10} + x_{10}^2 $
$c(3) = 0 = y_{10}^2 + x_{10}(x_{10}-2)(x_{10}-4) $
Both solutions of $c(1) = 0$ give $y_{10} = 0$. The contradiction proves the assertion of the OP.
Observations
In my first attempt I thought I needed the equality of the time period of the oscillation, which is equivalent to stating equality of the energies. But in fact it is not needed, as the developments above show.
If we would drop the term $x^2$ in the second equation we can have $d(t) = 0$ by taking $x_{10} = x_{20}$