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 

$s_{1}(0) = (x_{1}(t=0) = x_{10}, y_{1}(t=0) = y_{10})$     
$s_{2}(0) = (x_{2}(t=0) = x_{20}, y_{2}(t=0) = y_{20})$

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.

Observation: if we would drop the term $x^2$ in the second equation we can have $d(t) = 0$ by taking $x_{10} = x_{20}$