Suppose $a,b,c,d$ are consecutive prime numbers with $a<b<c<d$
Can $$a^b+b^c+c^d$$ be a prime number ?
On the one hand, I did not find a prime for $a\le 4723$ (maybe, someone double-checks this ?) , on the other hand, there are quartupels such that $a^b+b^c+c^d$ has no small prime factor. For example, the smallest prime factor of $$23^{29}+29^{31}+31^{37}$$ is $$1937815389893$$ having $13$ digits, the cofactor having $43$ digits, is prime as well. So, there is no easy prove that there is no such prime either.
The $n$-th number of the form you consider is of order $(n \log n)^{n \log n}= e^{n (\log n)^2 \log \log n}$.
The 'probability' that such a number is prime is $(n (\log n)^2 \log \log n)^{-1}$. The series $\sum (n (\log n)^2 \log \log n)^{-1}$ converges.
Thus, heuristically one expects only finitely many solutions. Since you did not yet find any among smaller numbers, chances are there are none.
I would guess to prove it is beyond current technology.
Given that there are examples without small factors it is also unlikely that a congruence condition can be found that excludes such numbers can exist.