How can I complete this proof by contradiction?

Yes, a proof by contradiction can be given. Suppose you have found a pair $(x,y)$ satisfying the equation.

Since $14$ and $2x^2$ are even, also $5y^2$ must be even as well. Therefore $y$ is even and so $y=2z$ for some integer $z$.

This implies $2x^2+20z^2=14$ that simplifies to $x^2+10z^2=7$. But $10z^2>7$ if $z\ne0$, so we must have $z=0$ and so $x^2=7$, a contradiction.

About your argument, there's a glitch. Since $14$ is even, two integers that sum to $14$ are either both even or both odd. However, since $2x^2$ is clearly even, you can conclude (as I did above) that also $5y^2$ must be even.

HINT: You’ll have a very hard time proving it this way. I recommend a different approach altogether. Note that $x^2$ and $y^2$ are non-negative, so $2x^2$ and $5y^2$ are at most $14$. Thus, if $y$ is an integer, then $y$ must be one of three integers; what are they? And what do they force $2x^2$ to be?

Added: And you can use your observation that $y$ must be even to reduce the possibilities still further.

Roundabout proof (not the one you should use :-) ):

$x^2 \equiv \{0,1,4,5,6,9\}\mod 10$

$\Rightarrow 2x^2 \equiv \{0,2,8,10,12,18\}\mod 20$

$y^2 \equiv \{0,1\}\mod 4$

$\Rightarrow 5y^2 \equiv \{0,5\}\mod 20$

$\Rightarrow 2x^2+5y^2 \equiv \{0,2,3,5,7,8,10,12,13,15,17,18\}\mod 20$

$\Rightarrow 2x^2+5y^2 \not\equiv 14\mod 20$

$\Rightarrow 2x^2+5y^2 \neq 14$