Try conceptual logic

we have, $x^k+y^(k+1)$ =k+2, now, $x^k+y^k$ = $x^k+y^(k+1)+y^k -y^(k+1)$ =k+2+ $y^k(1-y)$ so, 0<k<1, ∴ 2<k+2<3 now as y>k and both ∈ (0,1) so the value of $y^k$ will be small. suppose if y=0.5 and k=0.49 then $y^k$ <0.5 and (1-y)=0.5 so $y^k(1-y)$ <<0.5 so for any value, $y^k(1-y )$ atleast will not exceed to 0.5 and will not be less than 0 so 0< $y^k(1-y)$ <0.5 solving this we get, 2<k+2+ $y^k(1-y)$ <3.5

∴ 2< $x^k+y^k$ <3.5