WLOG $(a,b)=1$. We have $$(a-b)^2|a^2+b^2-(a-b)^2=2ab$$Also, $gcd((a-b)^2, a)= gcd((a-b)^2, b)=1$. Hence, $(a-b)^2|2$. Therefore, $a-b=\pm 1$. The rest follows.

Let $\gcd(a,b)=d$. Then, $\frac{(da)^2+(db)^2}{(da-db)^2} = \frac{a^2+b^2}{(a-b)^2}$ and $\frac{(da)^3+(db)^3}{(da-db)^3} = \frac{a^3+b^3}{(a-b)^3}$ so it suffices to solve the problem for $a,b$ such that $\gcd(a,b)=1$. Assume WLOG, $a>b$. Now, note that if $a-b\geq 2$ there exists a prime $p\mid a-b$. Then, \begin{align*} p & \mid a-b\\ p&\mid a^2-b^2\\ p& \mid a^2+b^2 \ \ \ \ \ \ \ \ \text{(by the assumed divisibility)}\\ p &\mid 2a^2 , 2b^2 \end{align*}If $p>2$ then, $p\mid a^2,b^2$ which implies that $p\mid a,b$ which is a contradiction to the fact that $\gcd(a,b)=1$. Thus, $p=2$. But then, $a,b$ must both be odd (if they are both even they have a common factor of 2). But we know, \[a^2 + b^2 \equiv 1 +1 \equiv 2\pmod{4} \text{ for all odd integers } a,b\]But clearly, $4\mid (a-b)^2$ which is a contradiction to the assumption that $\frac{a^2+b^2}{(a-b)^2}$ is an integer. Thus, our assumption must have been false and there exists no positive integers $a,b$ such that $\gcd(a,b)=1$, $a-b >1$ and $\frac{a^2+b^2}{(a-b)^2}$ is an integer. Thus, $\frac{a^2+b^2}{(a-b)^2}$ is an integer if and only if $a-b=1$ from which it is clear that $\frac{a^3+b^3}{(a-b)^3}$ is also clearly an integer. We are done.