I am a bit confused by the definition of irreducible Laurent polynomials. Given a Laurent polynomial, does it admit a unique factorization?
For instance, $2+x+x^{-1}$, obvious it is equal to $(x+1)(x^{-1}+1)=x(1+x^{-1})^2$. Does this suggest that factorization is not unique? Is $x+1$ irreducible?
The indeterminate $x$ is a unit (or, if you prefer, invertible element) in the ring of Laurent polynomials. Unique factorization in a ring doesn't distinguish between associates (i.e., pairs of elements $r, s$ such that $r = us$ for some unit $u$). E.g., $(-2)\times(-3)$ and $2 \times 3$ are not considered to be distinct factorizations of $6$ in $\Bbb{Z}$ (the units in $\Bbb{Z}$ being $1$ and $-1$). In your example: $$ x(1+x^{-1}) = 1 + x $$ so $1 + x$ and $1 + x^{-1}$ are associates and $(x+1)(x^{-1}+1)$ and $x(1+x^{-1})^2$ are thought of as two different ways of writing the same factorization into irreducibles. (And $x + 1$ certainly is irreducible.)