What are the eigenvalues of this Hamiltonian? \begin{equation}
H=E_{0} \sum_{m=-\infty}^{\infty}(|m\rangle\langle m+1|+\mathrm{h.c.})
\end{equation}
I tried to solve this equation as follows(h.c. means dagger or hermitian conjugate and m kets form orthonormal basis):
\begin{equation}
H|\psi\rangle=\lambda|\psi\rangle
\end{equation}
\begin{equation}
|\psi\rangle=\sum_{m^{\prime}=-\infty}^{\infty} \psi_{m^{\prime}}|{m^{\prime}}\rangle
\end{equation}
Which gives me:
\begin{equation}
\sum_{m^{\prime}} E_{0} \psi_{m^{\prime}}\left(\left|m^{\prime}-1\right\rangle+\left|m^{\prime}+1\right\rangle\right)=\sum_{m^{\prime}} \lambda \psi_{m^{\prime}}\left|m^{\prime}\right\rangle
\end{equation}
From which (please verify) I get:
\begin{equation}
\psi_{m^{\prime}+2}=\frac{\lambda}{E_{0}} \psi_{m^{\prime}+1}-\psi_{m^{\prime}}
\end{equation}
I could not advance here onwards.

*

*This recursion can grow from both sides which is fine, I assumed 0th and 1st psi to be some constants but it's without pattern, I am not getting how to proceed further. Help!

 A: Regarding your recursion relation: it is linear and thus may be solved by an ansatz of the form $\psi_m = a^{rm}$; $a, r$ unknown. We know that plane waves are going to feature$^\dagger$ in the solution so take $\psi_m = e^{ikm}$, Substituting in
$$e^{ikm}e^{2ik}=e^{ikm}e^{ik} \frac{\lambda}{E_0}-e^{ikm}$$
Dividing by $e^{ikm}e^{ik}$ and rearranging
$$\frac{\lambda}{E_0}= e^{ik}+e^{-ik}=2\cos(k)$$
A very general tip for such Hamiltonians: if they have terms such as $\left|n\right>\left<n+a\right|$, then they are likely diagonal in the Fourier transformed basis  $\left|k \right>=\sum_m e^{ikm} \left| m \right> $. So usually we would start with the plane wave eigenstates, rather than the general ones you used$^\ddagger$:
$$\left|\psi_k \right>=\sum_m e^{ikm} \left| m \right> $$
Applying $H$ to this, we find (after exploiting orthogonality, and $E_0=1$ for simplicity)
$$H \left|\psi_k \right>= \sum_m e^{ikm}\left( \left|m+1\right>+\left|m-1\right> \right)$$
Then relabeling summation indices
$$H \left|\psi_k \right>= \sum_m e^{ikm} \left( e^{ik}+e^{-ik}\right) \left|m\right> $$
$$H \left|\psi_k \right>=2\cos(k) \sum_m e^{ikm}\left|m\right> = 2\cos(k) \left|\psi_k\right>$$
From which we can read off the eigenvalues.
$\dagger$ We should really label these as $\psi_{k,m}$ or $\psi_m(k)$ or something similar, the eigenstates are labelled by $k$. Similarly $\lambda=\lambda(k)$.
$\ddagger$ To justify this, we could rewrite $H$ in the Fourier transformed basis, given by the inverse Fourier transforms
$$\left|m \right>=\sum_k e^{-ikm} \left| k \right> $$
And, taking the adjoint
$$\left<m \right|=\sum_k e^{ikm} \left< k \right| $$
Substitute these into $H$, you'll have three summations, two will be removed using orthogonality, and you should end up with exactly the same thing as with the plane wave ansatz.
