Bond energy and s-character
Let us start with the following most important part of the IUPAC definition for the bond energy.


  bond energy (mean bond energy)
  
  The average value of the gas-phase bond dissociation energies (usually
  at a temperature of 298 K) for all bonds of the same type within the
  same chemical species.


I think it should be immediately clear now that the higher a bond energy is, the stronger is the bond, so that there is no any contradiction in a bond energy increase with the increase of the $\mathrm{s}$-character in hybrid $\mathrm{sp}^n$-orbitals.

Indeed, $\mathrm{2s}$-orbitals are lower in energy that the $\mathrm{2p}$ ones, and thus, the more $\mathrm{s}$-character a hybrid orbital has, the lower is its energy: $E_{\mathrm{sp}} &lt; E_{\mathrm{sp^2}} &lt; E_{\mathrm{sp^3}}$. The lower an atomic or hybrid orbital energy is, the lower is the energy of a molecular orbital in formation of which it participates, $E_{\sigma_\mathrm{sp}} &lt; E_{\sigma_\mathrm{sp^2}} &lt; E_{\sigma_\mathrm{sp^3}}$, and consequently, the higher is the corresponding bond dissociation energy, as well as, the average of them, i.e. the bond energy. So, the bond energy indeed increases as the s-character grows, say $E_{\ce{C(\mathrm{sp})-C(\mathrm{sp})}} &gt; E_{\ce{C(\mathrm{sp^2})-C(\mathrm{sp^2})}} &gt; E_{\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}}$.

Below is the MO diagram for $\sigma$ $\ce{C-C}$ bond between two $\mathrm{sp}$-, two $\mathrm{sp^2}$-, and two $\mathrm{sp^3}$-hydridized $\ce{C}$ atoms that illustrates the logic above.