I have to find all positive integers possible for $x$ and $y$ that fit $2x+1$ is divisible by $y$ and $2y +1$ is divisible by $x$ I think I have to simultaneously solve by doing
$$2x + 1 = ay$$
$$2y + 1 = bx$$
But if I continue solving simultaneously, it kind of loops and I get nowhere.
 A: If $x=1$ you have $y \in \{1,3\}$. Otherwise without loss of generality $2\le x\le y$. Since $y\mid 2x+1$, then 
$$
x\le y\le 2x+1.
$$
Since $y$ is a divisor of $2x+1$, it can be only $2x+1$ or $\frac{2x+1}{d}$ with $d\ge 3$ odd (because the numerator is odd). The second case is impossible since
$$
x \le y =\frac{2x+1}{d} \le \frac{2x+1}{3} \Leftrightarrow x\le 1. 
$$
Hence $y=2x+1$. You miss only to verify that $x\mid 2y+1=4x+3$, which implies $x \in \{1,3\}$. The unique case to check is $x=3$ which leads to $y=7$. Therefore the unique solutions are
$$
\{(1,1),(3,1),(1,3), (3,7), (7,3)\}.
$$
A: We may assume $2x+1=my$, $\>2y+1=nx$ with $m\geq n\geq1$. Solving for $x$ and $y$ then gives
$$x={m+2\over mn-4},\qquad y={n+2\over mn-4}\ .$$
This implies $1\leq mn-4\leq n+2$, so that
$$m\geq3,\qquad (m-1)n\leq 6\ .$$
I this way the cases
$$\eqalign{n=1:\qquad&3\leq m\leq7 \cr n=2:\qquad&3\leq m\leq 6\cr n=3:\qquad&m=3\cr}$$
remain. For each of these test whether $x$ and $y$ become integer, and don't forget to incorporate also the mirror image of each found pair $(x,y)$ in your set of solutions.
A: Setting
$$2x+1=my$$ and $$2x+1=nx$$ then you will get
$$x=\frac{1}{2}(my-1)$$ and from here
$$2y+1=\frac{n}{2}(my-1)$$ solving for $y$ we have
$$y=\frac{n+2}{mn-4}$$ and you must use that $y$ is an intger number
