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One of the possible ordered triplets is (2,2,5).

However I am unable to disprove that there will never be another triplet of prime numbers satisfying that condition. something like (x,y,z) = (2,y,z). In this case, we will have the final result of the expressions as odd on both the sides of the equality.

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When $z$ is even, the only solution is when $z=2$, or $x^y=1$. This can only be possible if $x=1$ or $y=0$, so there is no solution.

When $z$ is odd, $x^y$ must be even, so $x$ must be even, which means $x=2$. However by this post, if $2^y+1$ is prime, $y$ must be a power of $2$. The only $y$ that satisfies this is $2$, so $(2,2,5)$ is the only solution.

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    $\begingroup$ A parallel observation (implicit in your reference to another post, but more generally true). If $y\gt 1$ is odd then the left-hand side has a factor $x+1$ and cannot be prime, so we must have $y=2$. $\endgroup$ Commented Sep 28, 2018 at 9:30
  • $\begingroup$ @MarkBennet Great observation! $\endgroup$
    – Toby Mak
    Commented Sep 28, 2018 at 9:32

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