If x = 2 , we have x^{2 }+ 2^{x} = 8 ( scrap because 8 isn't a prime number)

If x = 3 , we have x^{2} + 2^{x }= 17 ( satisfy )

If x > 3 and x is a prime number then x =3k + 1 or 3k + 2 ( k is a natural number)

+) x = 3k +1

If x = 3k +1 then x^{2 }divide to 3 has the remainder 1

and 2^{x} divide to 3 has the remainder 2

=> 2^{x }+ x^{2 }divided to 3

But 2^{x }+ x^{2 }> 3

So 2^{x }+ x^{2} isn't a prime number

Do the similar to x = 3k+2

The answer is x = 3