This statement is true for $E_n = 2$, since $E_{n+1} = 4 \cdot 2 + 0 = 8$. I was trying to take into account that there are always only two numbers between an even number and the next even number in the Fibonacci sequence. However, I didn't succeed.
That $E_{n+1}$ is an even number is an obvious fact, because $E_{n}$ and $E_{n-1}$ are both even. So I also tried to proof that there are no another even number between $E_n$ and $E_{n+1}$ in the Fibonacci Sequence.
The Fibonacci numbers mod $2$ are $0, 1, 1, 0, 1, 1, \ldots$. Thus the $n$'th even Fibonacci number is $F_{3n}$, and you're trying to prove $$ F_{3n+3} = 4 F_{3n} + F_{3n-3}$$ In fact it turns out that $$F_{k+3} = 4 F_k + F_{k-3}$$ for all $k$, not just those divisible by $3$. And this is easy to prove by induction: once you have it for $k$ and $k+1$, add two equations to get it for $k+2$.