Solution of this Diophantine Equation

The fault is that irrationals can also produce the product to $1$.
Consider $x=3$ and $y=2$ then we get, $(3+\sqrt{2} \cdot 2)(3-\sqrt{2}\cdot 2)=1$
Hence, the fault is moving from step 2 to step 3. You should look under the ring of $a+b\sqrt{2}$ in that step.

What about

\begin{align*}&x^2-2y^2=1\tag{1}\\\iff & x^2-1=(x+1)(x-1)=2y^2\end{align*}

Since $2\mid (x+1)(x-1)$, we conclude that both $(x+1)$ and $(x-1)$ have to be even, and hence $$4\mid 2y^2\implies 2\mid y^2\implies 2\mid y$$ and since $y$ is prime, $\color{red}{y=2}$. Can you end it now?

From (1), it follows immediately that $x^2-2y^2=x^2-8=1$. Thus, the only solution is $\color{blue}{(3, 2)}$.

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Addendum

The problem with your method is that for $a,b\in\mathbb R$

$$a·b=1\not\Rightarrow a=1\;\text{ and }\;b=1$$

In fact, this only works if $$a·b=0\implies a=0\;\text{ or }\;b=0$$