we will find the solution for x and y positive integer numbers

since x + y = 5 we have the following possibilities

x=1, y =4

1^1 + 4^4 = 257 <> 31

x=2, y=3

2^2 + 3^3 = 31 thus x=2, y=3 is a solution

x=3, y =2

3^3 + 2^2 = 31 thus x=3, y=2 is a solution

x = 4, y=1

4^4 + 1^1 = 257 <> 31

the positive integer solutions of the system

x^x+y^y=31

x+y=5

are (2,3) and (3,2).

you can also check the graph of the function

f(x) = x^x + (x-5)^(x-5) - 31

and notice that it has x-intercepts at x=2 and x=3

click here to see the graph

Click to see all the steps