We will prove this by contradiction. Denote the sum $x^c + y^d$ by $s$. Then, obviously, $x^c = -y^d \pmod{s}$ and also if $(x^a + y^b):s$, then $x^a = -y^b \pmod{s}$. This implies that $x^{ad} = (-1)^d y^{bd} \pmod{s}$ and $x^{bc} = (-1)^b y^{bd} \pmod{s}$, hence $(-1)^d x^{ad} = y^{bd} = (-1)^b x^{bc} \pmod{s} \Rightarrow x^{ad} = (-1)^{b-d} x^{bc} \pmod{s}$. It is clear that $x$ and $s$ are coprime, so we can divide the last congruence by $x$ raised to the power $\min\{ad, bc\}$, and obtain $x^{\max\{ad, bc\}} = (-1)^{b-d} \pmod{s}$. Similarly, we get $y^{\max\{ad, bc\}} = (-1)^{a-c} \pmod{s}$. Therefore, $y^{\max\{ad-bc\}} - x^{\max\{ad-bc\}}:s$ or $y^{\max\{ad-bc\}} + x^{\max\{ad-bc\}}:s$. By the statement of the problem, we have $0 < |ad-bc| < \min\{c, d\}$. But then
$$|y^{\max\{ad-bc\}} - x^{\max\{ad-bc\}}| < y^{\max\{ad-bc\}} + x^{\max\{ad-bc\}} < y^d + x^c = s.$$
So, the expression $|y^{\max\{ad-bc\}} \pm x^{\max\{ad-bc\}}|$ can be divisible by $s$ only if it is equal to zero. But this contradicts the fact that $x$ and $y$ are coprime, and we are done.