${a^2}{\mid}({b^3} + 1)\;$and$\;{b^2}{\mid}({a^3} + 1)$

Question

Prove that there are no integers $a,b \gt 2$ such that $a^2{\mid}(b^3 + 1)$ and $b^2{\mid}(a^3 + 1)$.

Answer 1

The counterexample $a=2,\,b=3$ ($4|28,\,9|9$) proves this isn't true.

Addendum for the new problem:

We have $a^2b^2|a^3+b^3+1$ and can replace $|$ with $\le$. Then $(a^2-b)(b^2-a)=a^2b^2-a^3-b^3+ab\le ab+1$, and without loss of generality $a\le b$ so $a^2\le b^2\le a^3+1< a^4$ implies $a^2>b$. Thus we have factorised a positive integer that is $\le ab+1$, each factor being positive. Then $b^2-a\le ab+1$ and $0\le b(b-a)\le a+1\le b+1$, and $b-a<1+\frac{1}{b}<2$. Hence $a=b$ or $a=b-1$. My guess is cheking these cases in turn shows we can't solve with $a\ge 3$.