Proving $a\mid b^2,\,b^2\mid a^3,\,a^3\mid b^4,\ldots\implies a=b$ - why is my approach incorrect

Hint $\ b(b/a),\,b(b/a)^3,b(b/a)^5\ldots\,$ are all integers, i.e. $\,b\,$ is a common denominator for $\rm\color{#c00}{unbounded}$ powers of $\,b/a,\,$ so $\,b/a,$ is an integer (prove it!), thus $\,a\mid b.\,$ Similarly $\,b\mid a.\,$

Hint $2\!:$ $\ $ Let $\, r = \frac{b}a = \frac{c}d,\ \color{#0a0}{(c,d)=1}.\,$ Then $\,b\:\!r^k = n_k\in\Bbb Z\,\Rightarrow\, b\:\!c^k = n_k d^k\,$ hence we infer by $\rm\color{#0a0}{Euclid}$: $\,d^k\mid b c^k\Rightarrow\, d^k\mid b^{\phantom{|^{|^|}}}\!\!$ for $\,\rm\color{#c00}{unbounded}$ $\,k\,$ so $\,d=\pm1,\,$ so $\,r = \frac{c}d = \pm c\in\Bbb Z,\,$ so $\,a\mid b$.

Remark $ $ That unbounded powers of proper fractions have no common denominator (such as $b$ above), is a special property of $\,\Bbb Z\,$ that needn't be true in general domains. When it holds true, a domain is called completely integrally closed.

See this answer for a generalization to Noetherian integrally closed domains, e.g. PIDs.