Ignoring the axiom of regularity (and therefore the implication of "no set can contain itself"), would it be correct to state that the set that contains only itself is unique?
My argument is that if $x$ is said set, then $$ x = \{x\} = \{\{x\}\} = \{\{\{\cdot\cdot\cdot\}\}\} $$ Ad infinitum, which seems to be unique.
On
It need not be unique; in fact, you can weaken the axiom of foundation to allow either well-founded sets or sets of the form $x=\{x\}$. Sets of the latter form are called Quine atoms, and play the role of urelements. These are useful in set theory with atoms because they allow you to formulate a theory of sets-and-atoms without resorting to multi-sorted logic.
For more, see The Axiom of Choice by T. J. Jech.
In Vicious Circles, Barwise and Moss explore circularities like these that can result when the ZF Axiom of Foundation is dropped. In their presentation, uniqueness of solution of the equation $x=\{x\}$ and others like it must be added as an axiom.