Defining the $2 \times 2$ transfer matrix
\begin{align}\tag{1}
Q = \begin{pmatrix} -\lambda & 1 \\ -1 & 0 \end{pmatrix},
\end{align}
the characteristic polynomial (CP) of the $M \times M$ matrix $A_M$ is given by
\begin{align}
P_M(\lambda) &= \langle 1{-}\lambda, 2| \, Q^{M-1} \,|1,0\rangle \tag{2a}\\
&= 2 T_M(-\lambda/2) + U_{M-1}(-\lambda/2) \tag{2b}\\
&= 2\cos(M \varphi) + \frac{\sin(M \varphi)}{\sin(\varphi)}\tag{2c},
\end{align}
with Chebyshev polynomials $T_M,U_M$, and with $\lambda=-2\cos\varphi$.

The unnormalized right eigenvectors $A_M x_\mu=\lambda_\mu x_\mu$ have the elements
\begin{align}
x_{\mu,m} = \langle 1, 0| \, (-Q)^m \, |1,0\rangle,\quad m=0,\ldots,M-1.\tag{3}
\end{align}
The eigenvector normalization can be related to the derivative $P_M'(\lambda)$, see, e.g., https://arxiv.org/abs/2103.10776 for details.

Regarding to your question, I don't think that a closed form expression exist for $\lambda_\mu$ if $M>5$, as due to the left boundary term, the CP does not factorize in this case (up to one trivial eigenvalue $\lambda=\pm1$ if $M=3n\pm1$). However, it is often not necessary to explicitly calculate the eigenvalues, see my cited paper, as the CP together with the eigenvectors contains enough, or even more, information.