Consider an absolutely incompressible body, which does not change its volume $V$ under pressure $p$ but can expand or shrink when its temperature $T$ changes. Then its equation of state has a simple form $$ T=T(V). $$
Then imagine we are performing the Brayton cycle with this body consisting of two isobaric processes at the pressures $p_1$ and $p_2$, and two adiabatic processes. The latter two processes are also isothermal and isochoric since the body is incompressible.
The efficiency of this cycle is $\eta=A/Q_1$, where $A=(p_1-p_2)(V_1-V_2)$ is the performed work and $Q_1$ is the heat received from the hot reservoir. To calculate $Q_1$, we can use the first law of thermodynamics: $\delta Q=c_VdT+pdV$. The Maxwell's relations imply the infinite heat capacity ratio $\gamma=c_p/c_V\rightarrow\infty$ for incompressible body then $c_V=0$ (it can be also understood considering that $T$ and hence the internal energy do not change at $V=\mathrm{const}$).
So $\delta Q=pdV$, and the heat, which is revieved by the system at the top isobare, is $Q_1=p_1(V_1-V_2)$. The efficiency becomes $$ \eta=1-\frac{p_2}{p_1}. $$
From the other side, the Carnot efficiency $\eta_\mathrm{Carnot}$ depends on the ratio of the highest $T_1=T(V_1)$ to the lowest $T_2=T(V_2)$ temperatures during the cycle, and $$ \eta_\mathrm{Carnot}=1-\frac{T(V_2)}{T(V_1)}. $$ Taking the pressure difference $p_1-p_2$ large enough while keeping $V_1-V_2$ constant, we can achieve $\eta>\eta_\mathrm{Carnot}$.
Why the Carnot's theorem is violated in this case? Does it mean that the Carnot's theorem prohibits existence of absolutely incompressible materials? Or something in my calculations is wrong?
UPDATE
Indeed, as noted in the comments, the Carnot's theorem is applicable when only two heat reservoirs with fixed temperatures are present. However, it can be easily generalized to the case of variable temperatures. If $T_1$ is the maximum temperature achieved by the reservoir which transfers heat to the system and $T_2$ is the minimum temperature of the second reservoir which absorbs heat then the efficiency of any cycle cannot exceed $$ \eta_\mathrm{max}=1-\frac{T_2}{T_1} $$ (see, e.g., R. Kubo, Thermodynamics (1968), problem 4 on page 85 and its solution on pp. 97-98).
Moreover, I have found even stranger thing when applied the proof of this relation to incompressible body. Assume "a" and "b" are, respectively, the top and bottom isobares where the system absorbs and gives off amounts of heat $Q_1$ and $Q_2$. Then, according to the Clausius' inequality, $$ \int\limits_a\frac{\delta Q_1}{T}\leqslant\int\limits_b\frac{\delta Q_2}{T}. $$ As stated above, for incompressible body $\delta Q=pdV$ then $$ \int\limits_a\frac{\delta Q_1}{T}=\int\limits_{V_2}^{V_1}\frac{p_1dV}{T(V)}, \qquad\int\limits_b\frac{\delta Q_2}{T}=\int\limits_{V_2}^{V_1}\frac{p_2dV}{T(V)}. $$ Canceling $\int_{V_2}^{V_1}dV/T(V)$ on both sides of the inequality, we get $$ p_1\leqslant p_2 $$ in contradiction with the initial assumption $p_1>p_2$ about the shape of the cycle.
So, now I suppose that even the Clausius' inequality prohibits existence of incompressible bodies. Is this true and how it can be understood?
UPDATE NUMBER 2
I tried to state the problem anew. Assume the system at given state variables $T$ and $p$, it is characterized by the Gibbs potential $G$ with $$ dG=-SdT+Vdp,\qquad\left(\frac{\partial G}{\partial p}\right)_T=V,\qquad\left(\frac{\partial G}{\partial T}\right)_p=-S, $$ so incompressibility $\partial V/\partial p=0$ means $V$ depends only on $T$: $(\partial G/\partial p)_T=V(T)$. Integrating this relation, we get $$ G=pV(T)+A(T), $$ where $A(T)$ is some unknown function. Accordingly, we get the entropy $$ S=-pV'(T)-A'(T) $$ and the internal energy $U=G+TS-pV$: $$ U=-pTV'(T)-TA'(T)+A(T). $$
However, it this case the calculations of the cycle efficiency should be reconsidered because $$ \delta Q=TdS=-TV'dp-(pTV''+TA'')dT $$ implies heat exchange during isothermal/isochoric processes! Thus my assumption about adiabatic nature of isochoric processes is incorrect.
For example, assuming $V=\alpha T$, $A=0$, we get $\Delta Q=0$ at the both isobares, $\Delta Q=(p_1-p_2)T_1\alpha=Q_1$ at the right isochore and $\Delta Q=-(p_1-p_2)T_2\alpha=-Q_2$ at the left isochore. Thus the efficiency $$ \eta=1-\frac{Q_2}{Q_1}=1-\frac{T_2}{T_1} $$ matches the Carnot's limit.
So what is wrong in this case? I see that some results seem strange (e.g., dependence of $U$ on $p$ at $T=\mathrm{const}$), but are there any fundamental problems (with stability, existence of thermodynamic equilibrium etc.)?