Let $n=250\cdot 3^{6k}$ ($k \in \mathbb{N}$) We can verify that $n$ is a sum of two squares and a sum of two cubes as follows: $n = (5\cdot3^{3n})^2 + (15\cdot3^{3n})^2$ $n = (5\cdot3^{2n})^3 + (5\cdot3^{2n})^3$ Assume that there exists $(a,b)\in \mathbb{Z}^2$ for which $a^6+b^6=n$. Let $(d,s,t)$ be a triplets of integers for which $\gcd(a,b)=d$ and $(a,b)=(ds,dt)$. Then we have $d^6(s^6+t^6)=250\cdot 3^{6k}$ Since $3 \nmid s^6+t^6$, we have $3^{6k} \mid d^6$, which implies $d/3^k \in \mathbb{Z}$. So we have $(d/3^k)^6\cdot (s^6+t^6)=250$. Since there does not exist a prime $p$ for which $p^6 \mid 250$, we must have $d/3^k=1$. Finally, we have $s^6+t^6=250$. Since $250<3^6$, we have $|s|,|t| \leq 2$ implying $250 \leq 2^6+2^6=128$. Thus we have reached a contradiction.

utkarshgupta wrote: Show that there are infinitely many natural numbers which are simultaneously a sum of two squares and a sum of two cubes but which are not a sum of two $6-$th powers. We claim that if $n$ satisfies then $64n$ satisfies as well. This is beacuse, if $n=a^2+b^2=c^3+d^3$ then, $64n=(8a)^2+(8b)^2=(4c)^3+(4d)^3$ and if, $64n = x^6+y^6$ then $x,y$ have to be even (check mod 4). Which will imply $n$ can be written as sum of two 6th powers. A contradiction. Therefore it suffices to find one solution. To restrict our $n$ to not be sum of two 6th powers, we should try modular restriction. Modulo 7 is nice because of FLT. By some hit and trial $370=7^3+3^3=3^2+19^2$ and $370 \equiv 6 \pmod 7$.