Let $f(z) = z^2 + c$ for some $c\in\Bbb C$, and let $R=\dfrac{1+\sqrt{1+4\lvert c\rvert}}2$. If for some $a\in\Bbb C$ and $n>0$ we have that $\lvert f^n(a)\rvert>R$, then $f^n(a)\to\infty$ as $n\to\infty$ — i.e., $a$ is in the basin of attraction to infinity, so it is not in the closed Julia set.
There is a lot going on there. What I would like to know is whence this value of $R$ came and why the threshold of attraction to infinity can be simply confined to some circle.
I would try to do more research on my own, but as it was presented to me this theorem is nameless.
Christoph Stroh in his thesis "Julia Sets of Complex Polynomials and Their Implementation on the Computer" proves that the Julia set of $f$ is contained in the disk of radius $R$ centered at the origin. See section 3.1, especially example 3.8.
The expression for $R$ comes from the largest fixed point of $f$: $$ z^2+c =z \iff z = \frac{1 - \sqrt{1 - 4 c}}{2} $$ This fixed point certainly belongs to the Julia set. Actually, this point is on the border of the convex hull of the Julia set.
The value given for $R$ allows us to prove that $|f(z)| > R$ if $|z|>R$: $$ |f (z)| = |z^2 +c| \ge |z^2|−|c| = |z|^2 −|c| > |z| > R $$ because $|z|^2 −|c| > |z|$ if $|z| > R$.
But this is not enough to prove that $|f^n(a)| \to \infty$. More later.