Tridiagonal shifts as compact + isometry

Let $\{a_n\}_{n\geq 0}$ and $\{b_n\}_{n\geq 0}$ be sequences of scalars. Suppose $a_n \neq 0$ for all $n \geq 0$. We consider the tridiagonal kernel (also known as band kernel with bandwidth one) as $$k(z, w) = \sum_{n=0}^\infty ((a_n + b_n z)z^n) \overline{(({a}_n + {b}_n {w}) {w}^n)} \qquad (z, w \in \mathbb{D}),$$ where $\mathbb{D} = \{z \in \mathbb{C}: |z| < 1\}$. Denote by $M_z$ the multiplication operator on the reproducing kernel Hilbert space corresponding to the kernel $k$. Assume that $M_z$ is left-invertible. We prove that $M_z =$ compact $+$ isometry if and only if $|\frac{b_n}{a_n}-\frac{b_{n+1}}{a_{n+1}}|\rightarrow 0$ and $|\frac{a_n}{a_{n+1}}| \rightarrow 1$.