The energies and states you get are correct. I will point out something else that may be useful. I will use the following basis: $$|00> = |0>;\quad|01> = d^\dagger|0>; |10> = f^\dagger|0>;\quad |11>=f^\dagger d^\dagger |0>.$$ The Hamiltonian at hand conserves parity, so you can write the Fock-space matrix in a more convenient way: $$H=\begin{array}{c|cc|cc} \,&|00>&|11>& |01> & |10> \ \hline <00| & -(\epsilon_m + \epsilon_d/2) & t & 0 & 0\ <11| & t & (\epsilon_m + \epsilon_d/2) & 0 & 0 \ \hline <01|&0 & 0& -(\epsilon_m-\epsilon_d/2) & t \ <10| & 0 & 0 & t & \epsilon_m-\epsilon_d/2 \end{array},$$ which is block-diagonal, so you only have to deal with subspaces of size $2\times 2$. Let's label the parities of the subspaces as $P=0$ and $P=1$, respectively, as $P=(n_f+n_d)\text{mod}2$. The eigenvalues are $$E_{0,\pm}=\pm\frac{\sqrt{(\epsilon_d+2\epsilon_m)^2+4t^2}}{2};\quad E_{1,\pm}=\pm\frac{\sqrt{(\epsilon_d-2\epsilon_m)^2+4t^2}}{2},$$ where the index $0$ or $1$ refers to the parity. 526
You already got the normalized eigenstates, and all you need to do is reorganize them to put them in this particular ordering. To make things quicker, I'll simply write them as follows: $$|0\pm>=\alpha_0^{\pm}|00>+\beta_0^{\pm}|11>;\quad |1\pm>=\alpha_1^{\pm}|01>+\beta_1^{\pm}|10>,$$ with $\alpha_P^{\pm}$ and $\beta_{P}^{\pm}$ the coefficients you found. You can associate to these the operators $A_{0,\pm}^\dagger|0> = |0\pm>$ and $A_{1,\pm}^\dagger|0>=|1\pm>$.
Now, the operators $A_{1,\pm}$ have fermion commutation relations, because they are single-particle operators: $${A_{1,\sigma}^\dagger,A_{1,\sigma'}}={\alpha_{1,\sigma}d^\dagger+\beta_{1,\sigma}f^\dagger,\alpha_{1,\sigma'}^d+\beta_{1,\sigma'}^f}=\alpha_{1,\sigma}\alpha_{1,\sigma'}^+\beta_{1,\sigma}\beta_{1,\sigma'}^=\delta_{\sigma,\sigma'},$$ by definition (the coefficients meet this condition, because they are the coefficients of mutually orthonormal vectors). For the $P=0$ subspace, what you have are actually bosons, but not normal bosons, because they mix different parts of Fock space ($0$ particle and $2$ particle). As a consequence, you cannot get normal boson commutation relations: $$\begin{split}[A_{0,\sigma}^\dagger,A_{0,\sigma'}]=&[\alpha_{0,\sigma}+\beta_{0,\sigma}f^\dagger d^\dagger,\alpha_{0,\sigma'}^+\beta_{0,\sigma'}^d f]\ =& \beta_{0\sigma}\beta_{0\sigma'}^*\left(d^\dagger d + f^\dagger f -1 \right). 66