<s>[INST] Solution of this Diophantine Equation 
If $x$ and $y$ are prime numbers which satisfy $x^2-2y^2=1$, solve for $x$ and $y$.

My attempt: 
$x^2-2y^2=1$
$\implies (x+\sqrt{2}y)(x-\sqrt{2}y)=1$
$\implies (x+\sqrt{2}y)=1$ and $(x-\sqrt{2}y)=1$
$\implies x=1$ and $y=0$
Clearly $x$ and $y$ are not prime numbers . Why is my solution not working. I have been able to solve similar type of equations by factorizing and then listing down the integer factors and the different cases. Why is it not working here?
 [/INST] 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 $\color{blue}{(3, 2)}$.

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 $$a·b=0\implies a=0\;\text{ or }\;b=0$$
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