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Nash's theorem says that every finite game, including a two-player game with two pure choices each, has at least one Nash equilibrium in mixed strategies.

For player $1$, holding $p_2$ fixed, the expected value of the payoff isIt is an affine function of $p_1$. Therefore player $1$ chooses $p_1=1$ when $1-(U-1)p_2>0$, chooses $p_1=0$ when it is negative, and is indifferent when it vanishes. By symmetry, player $2$ has the analogous best response.

If $U<2$, then $1-(U-1)p>0$ for every $p\in [0,1]$, so becoming a Dark Lord with probability one is a strictly dominant strategy. The unique equilibrium isIf $U>2$, each player is indifferent when the other chooses $p=1/(U-1)$. This gives one interior mixed equilibrium, while the two asymmetric pure equilibria arise because a player facing a certain peasant chooses Lord and a player facing a certain Lord chooses peasant. The three equilibria areConsequently $\boxed{U_0=2}$. At the threshold itself there is a continuum of equilibria with $p_1=1$ or $p_2=1$, consistent with the question's strict inequalities.