If you're formulating a "set theory" and you allow sets to arbitrarily* self-reference, then you end up with Russell's paradox. (*I just learned today about Quine atoms. While they are not allowed in ZF, they allow for self-reference and don't have to allow for Russell's paradox, so perhaps self-reference is okay, sometimes.)
A solution is the axiom of foundation, which forbids such self-referential sets.
Now consider this:
The Gödel sentence is designed to refer, indirectly, to itself. The sentence states that, when a particular sequence of steps is used to construct another sentence, that constructed sentence will not be provable in F. However, the sequence of steps is such that the constructed sentence turns out to be GF itself. In this way, the Gödel sentence GF indirectly states its own unprovability within F (Smith 2007, p. 135).
In the case of Russell's paradox, adding an axiom to disallow self-reference prevented the sentence from being formed.
Similarly, is there an axiom we can introduce that disallows Gödel sentences from being formed?
No. Adding axioms can never prevent a construction from being carried out.
On a related note, you can never resolve a contradiction by adding more axioms. (in fact, if you have a contradiction in a theory, any new axioms you might consider adding are already theorems, by the principle of explosion!)
The key element to resolving Russell's paradox is not adding the axiom of foundation — it is removing the axiom of unrestricted comprehension, and replacing it with the more limited axiom of subsets.
Russell's paradox relies on having the axiom of unrestricted comprehension which implies that $\{ x \mid x \notin x \}$ is a set. If you add the axiom of foundation... the axiom of unrestricted comprehension still implies that $\{ x \mid x \notin x \}$ is a set.
But if you remove the axiom of unrestricted comprehension? In ZFC, there is no known proof that $\{ x \mid x \notin x \}$ is a set. And therefore, there is no known way to turn Russell's argument (which proves $\{ x \mid x \notin x \}$ is not a set) into a proof that ZFC leads to contradiction.