Proof by induction; $a^n$ divides $b^n$ implies $a$ divides $b$

Let $a,b$ be integers and let $P(k)$ be the statement "$a^k \mid b^k$ implies $a \mid b$." You are trying to prove by induction that $P(k)$ is true for all $k\in \mathbb{N}$.

You start out by proving the base case $P(1)$, which is fine. Then you say "suppose this is true for some $k$," by which I mean you are supposing that $P(k)$ is true. But then you say, "then $a^k \mid b^k$." This does not follow from $P(k)$, since $P(k)$ says nothing about whether $a^k$ divides $b^k$ or not, just about what happens if it does.

What you need to do is show that if $P(k)$ is true, then $P(k+1)$ is also true. Others have already talked about how to approach this, I just wanted to clarify a bit more what goes wrong with your attempt.

That doesn't quite work. What you need to do is assume that for some $k\geq 1$ we have $a^k\mid b^k$ implies $a\mid b$ for all integers $a,b$. Now, suppose that $a^{k+1}\mid b^{k+1}$ for some $a,b$, and show that $a\mid b$. Given the inductive hypothesis, it will be enough for you to show that $a^k\mid b^k$.

Inductive step should be $\,(P_k \Rightarrow Q)\Longrightarrow (P_{k+1}\Rightarrow Q),\,$ where $\,P_k := a^k\mid b^k,\ $ $\,Q := a\mid b$

But you proved $\ \ (P_k\ \,\&\,\ (P_k \Rightarrow Q)) \Longrightarrow\, P_{k+1},\,$ which is not equivalent. Below are correct ways.

Hint $\, $ By homogeneous reduction (below) we may assume $\rm\,a,b\,$ coprime. Then it suffices to show that $\rm\,a\mid b^n\,\Rightarrow\,a\mid 1\,$ (so $\rm\,a\mid b).\,$ By Euclid's Lemma $\rm\ a\mid b(b^k)\,\Rightarrow\,a\mid b^k,\,$ so $\rm\, a\mid 1\,$ by induction.

Remark $\, $ Alternatively if $\rm\,(a,b)=1,\,$ and $\rm\,b^n/a^n = m\in\Bbb Z\,$ then $\rm\rm\, b/a\,$ is a rational root of $\rm\,x^n - m,\,$ so, by RRT = Rational Root Test, $\rm\,b/a\in \Bbb Z,\,$ i.e. $\rm\,a\mid b.$

The first proof essentially uses an inductive extension of Euclid's Lemma: $ $ if $\rm\ a\ $ is coprime to all $\rm\,b_{\:\! i}\,$ then it is coprime to their product (more conceptually: units (invertibles) $\rm\,b_{\:\! i}\bmod{a}\,$ are closed under multiplication). This is the key lemma at the core of the proof of RRT, so it too may be considered an inductive proof, since it invokes said inductive extension of Euclid's Lemma. If that doesn't satisfy whatever fuzzy notion of "true inductive proof" one employs, then this proof likely will. It is a little-known proof of monic RRT by induction on degree. As a particular case, one can prove cube-roots irrational by using irrationality proofs of square-roots!

Homogeneous reduction to the case $\rm\,\color{#c00}{a,b\ \text{coprime}}$ works more precisely as follows. Suppose that $\rm\,c = (A,B).\,$ Then $\rm\, A,B = ac,bc,\,$ where $\rm\,(a,b) = 1,\,$ hence

$$\rm A^n\mid B^n\Rightarrow\, a^n c^n \mid b^n c^n\Rightarrow \color{#c00}{a^n\mid b^n \Rightarrow a\mid b} \Rightarrow\, ac\mid bc\Rightarrow A\mid B$$

Thus the noncoprime case $\rm\, A^n\mid B^n\Rightarrow A\mid B\,$ follows from the coprime case $\rm\color{#c00}{a^n\mid b^n \Rightarrow a\mid b}$