Yesterday, at 23:18, I thought it was a remarkable moment of the day. The digits on the watch were providing a quadruplet of positive integers that satisfy the following system of equations: $$\begin{align} a^b &= cd \\ b^a &= c+d \end{align}$$
I wondered what the set of all positive integer solutions of this system was. Writing $d=b^a-c,$ I obtained a quadratic of $c$ with coefficients in terms of $a$ and $b$, which led me to the following: $$\left\{ c,d \right\} = \left\{\frac{b^a-\sqrt{b^{2a}-4a^b}}{2},\frac{b^a+\sqrt{b^{2a}-4a^b}}{2}\right\}$$
Therefore, given positive integers $a$ and $b,$ there is a solution if and only if $b^{2a}-4a^b$ is a perfect square. A brute force search using this result yields the following solutions: $(1,2,1,1),$ $(2,2,2,2),$ $(2,3,1,8),$ and $(2,3,8,1).$ I believe that there is no other solution than these. However, I'm unable to prove it. I would be glad if anyone could help me.