Solve the following equation in integers $x,y:$ $x^2+6xy+8y^2+3x+6y=2.$ Question: Solve the following equation in integers $x,y:$ $$x^2+6xy+8y^2+3x+6y=2.$$
Solution: For some $x,y\in\mathbb{Z}$ $$x^2+6xy+8y^2+3x+6y=2\\\iff x^2+2xy+4xy+8y^2+3x+6y=2\\\iff x(x+2y)+4y(x+2y)+3(x+2y)=2\\\iff(x+4y+3)(x+2y)=2.$$ 
Now if $(x+4y+3)(x+2y)=2$, then either $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\text{ or }\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\text{ or }\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\text{ or }\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}.$$ 
We have $$\begin{cases} x+4y+3=1\\ x+2y=2\end{cases}\iff (x,y)=(6,-2), \\\begin{cases} x+4y+3=2\\ x+2y=1\end{cases}\iff (x,y)=(3,-1), \\\begin{cases} x+4y+3=-1\\ x+2y=-2\end{cases}\iff (x,y)=(0,-1),\\\begin{cases} x+4y+3=-2\\ x+2y=-1\end{cases}\iff (x,y)=(3,-2).$$ 
Now since, all the four pairs $(6,-2),(3,-1),(0,-1),(3,-2)$ satisfies the integer equation $(x+4y+3)(x+2y)=2$, thus we can conclude that $(x+4y+3)(x+2y)=2\iff (x,y)=(6,-2),(3,-1),(0,-1),(3,-2).$ 
Hence, we can conclude that the integer equation $x^2+6xy+8y^2+3x+6x=2$ is satisfied if and only if $(x,y)=(6,-2),(3,-1),(0,-1),(3,-2)$, and we are done. 
Is the solution correct and rigorous enough? And, I am always confused while solving equations regarding the usage of the if and only if arguments, which I feel is very necessary in order to have a complete and rigorous solution, but I rarely find it's usage in any book while solving equations of any kind. So, is it necessary? Also, is there a better solution than this? 

Another way:
Discriminant $$=(6y+3)^2-4(8y^2+6y-2)=4y^2+12y+17=(2y+3)^2+8$$ which needs to be perfect square $=z^2$(say)
WLOG $z\ge0$
so that $$x=\dfrac{-(6y+3)\pm z}2$$
Now $$(2y+3)^2-z^2=8$$
$$\iff\dfrac{2y+3-z}2\cdot\dfrac{2y+z+3}2=2=1\cdot2=(-2)(-1)$$
As $2y+3+z\ge2y+3-z,$
$\dfrac{2y+3+z}2=2\iff\dfrac{2y+3-z}2=-1$ 
Or $\dfrac{2y+3+z}2=-1$
