find all integer solutions for $xy+1=3(x+y)$ the original problem was to find all integer solutions for  $xy+1=3(x+y)$.
$$xy+1=3(x+y)$$
$$xy+1=3x+3y$$
$$xy-3x=3y-1$$
$$x(y-3)=3y-1$$
$$x=\frac{3y-1}{y-3}$$
I was able to find all possible solutions using this method,
I was wondering if you can say that $3|xy+1$ or $x+y|xy+1$ and proceed to solve it in this fashion.
If it is possible please show me how.

We want to find a factorization if possible, so consider $xy-3x-3y+k = k-1$ as a re-writing of the condition. Then we can take any value for $k$ to make a factorization possible, and $(x-3)(y-3) = xy-3k-3y+9$, so take $k=9$ to get 
$$(x-3)(y-3) = 8$$
Then the integer ordered factor pairs of $8$ are 
$$\{(-1,-8), (-2,-4), (-4,-2), (-8,-1),(1,8), (2,4), (4,2), (8,1)\}$$ and the possible solutions $$(x,y)\in\{(2,-5), (1,-1), (-1,1), (-5,2),(4,11), (5,7), (7,5), (11,4)\}$$

Looking at $(x+y)\mid(xy+1)$, it seems like the best result from that probably takes us back on the same track, ie
\begin{align}
j(x+y)&=xy+1 \\
xy-jx-jy+1 &= 0 \\
xy-jx-jy + j^2 &= j^2-1 \\
(x-j)(y-j) & = j^2-1
\end{align}
... except here of course we already know what $j$ is.
In a different problem,
$\begin{align}(x+y) &\mid(xy+1) \\ 
\Rightarrow(x+y) &\mid(xy+1)-(x+y) \\
\Rightarrow(x+y) &\mid(xy-x-y+1) \\
\Rightarrow (x+y) &\mid(x-1)(y-1)
\end{align}$
might be useful.
