Cute triplets

We know that $x^{y}+1^{y}=z=$ prime.Now,as long as $y=odd$ $x^{y}+1^{y}$ which is a prime is divisible by $x+1$ which is even (excluding $2$ ).Thus, $y=$ even.Now,the only possible case is when $y=2.$ $\Longrightarrow x^{2}+1=z.$ But as long as $x=$ odd and $z=$ odd,this can't happen.Thus,the only possible case is when $x=2.$ Putting $x=2$ in the first equation gives $z=5.$ Thus,the only possible triplet $x,y,z=2,2,5.$

The only ordered pair is $(2,2,5)$

Hint :

x,y, and z are prime numbers and prime numbers can be odd numbers only (except 2)

$\large (odd)^{odd}=odd .$

$\large odd+1=even=z (not possible)$ except 2

!!!

x,y, and z are prime numbers and prime numbers can be odd numbers only (except 2) answer is (2,2,5)