<s>[INST] 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.
 [/INST] 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
$$
\{(1,1),(3,1),(1,3), (3,7), (7,3)\}.
$$
</s>