In a lecture on Quantum mechanics, the professor concluded that if $a$ is a linear operator with $[a, a^\dagger] = 1$, where $a^\dagger$ is the adjoint of $a$ and $[a, a^\dagger] = aa^\dagger - a^\dagger a$, and if $a^\dagger a$ has an eigenvalue, then vectors of form \[ | n \rangle = (a^\dagger)^n | 0 \rangle \quad ( n \in \mathbb Z_{ \ge 0 }) \] are all eigenvectors of $a^\dagger a$. ($| \lambda \rangle$ is the eigenvector of $a^\dagger a$ which belongs to the eigenvalue $\lambda$).

But he gives no reason why its eigenvalue exists. So, I asked him if an eigenvalue exists, and he said 'I cannot answer. Roughly speaking, it is some limit of the case of a finite dimensional complex vector space.'

My question is:
Let $H$ be an infinite dimensional Hilbert space, say $\ell^2$ space, whose inner product is denoted by $\langle \cdot, \cdot \rangle$. It is well-known that for a bounded linear operator $T \in \mathcal L(H)$, there corresponds a (bounded) linear operator $T^\dagger \in \mathcal L(H)$ such that $\langle x, Ty \rangle = \langle T^\dagger x, y \rangle$ for all $x, y \in H$.
Then, is it true that *for any bounded linear operator $T \in \mathcal L(H)$ with $[T, T^\dagger] = 1$, there exists a scalar $\lambda \in \mathbb C$ and a nonzero vector $x \in H$ such that $T^\dagger T x = \lambda x$*?

How could I deal with it? Is there a 'characteristic polynomial' of $T^\dagger T$?

Please give me the answer, or some references.