It can be shown that the spectrum of a unitary operator from a Hilbert space to itself is a subset of the unit circle, as in $|\lambda|=1$
It can also be shown that the spectrum of the shift operator (more specifically the right shift operator from $\ell^2$ to $\ell^2$) contains points inside the unit circle
The first answer in this post also shows that the right shift operator is unitary (this is for $\ell^2(\mathbb{Z})$ but the right shift operator in $\ell^2(\mathbb{N})$ is also unitary, proof below)
What am I misunderstanding that is causing me to see this as contradiction? Is my definition of a unitary operator incorrect?
Proof that the right shift operator is unitary: an operator is unitary if $||Tx||=||x||$ and $\langle Tx,Ty\rangle=\langle x,y\rangle$
First property: $$||Tx||=\sum_{n=1}^\infty|(Tx)_n|^2=\sum_{n=2}^\infty|x_{n-1}|^2+0=\sum_{n=1}^\infty|x_n|^2=||x||$$
Second property: $$\langle Tx,Ty\rangle=\sum_{n=1}^\infty (Tx)_n(Ty)_n=\sum_{n=2}^\infty x_{n-1}y_{n-1}+0=\sum_{n=1}^\infty x_ny_n=\langle x,y\rangle$$
The part that probably slipped by you is that $\ell^2(\mathbb N) \ne\ell^2(\mathbb Z)$.
But to be complete we should note that while it is possible to give an easy isomorphism between the two spaces ($\mathbb Z$ and $\mathbb N$ are both countable infinite sets, and any bijection between them induces a Hilbert space isomorphism), the two right shift operators do not correspond to each other under such an isomorphism.
And what closes the book on this is noting that while the operator on $\ell^2(\mathbb Z)$ is actually unitary, $U = $ right shift on $\ell^2(\mathbb N)$ has $U^*$ = left shift, so $UU^* = $ "zero out the first component", so that operator is not unitary.