From $2x+3y=185$, we have $(x,y) = (-185+3k,185+2k)$ where $k\in \mathbb{Z}$. Because $xy>x+y$, we have $6k^2 + 180k -185^2>0$. Then, $$k > \frac{5\sqrt{8538}- 90}{6} \quad \text{or}\quad k<\frac{-5\sqrt{8538} - 90}{6}.$$ Because $k\in \mathbb{Z}$, =we conclude that $k\ge 62$ or $k\le -93$. So, the solutions are $$\boxed{(x,y) = (-185+3k,185+2k), \; k\in \mathbb{Z},\; k\in (-\infty,-93]\cup [62,\infty)}.$$

From $2x+3y=185$, we have $(x,y) = (-185+3k,185+2k)$ where $k\in \mathbb{Z}$. Because $xy>x+y$, we have $6k^2 + 180k -185^2>0$. Then, $$k > \frac{5\sqrt{8538}- 90}{6} \quad \text{or}\quad k<\frac{-5\sqrt{8538} - 90}{6}.$$ Because $k\in \mathbb{Z}$, =we conclude that $k\ge 62$ or $k\le -93$. So, the solutions are $$\boxed{(x,y) = (-185+3k,185+2k), \; k\in \mathbb{Z},\; k\in (-\infty,-93]\cup [62,\infty)}.$$

$2x+3y=185$ Integer solutions: $y=1+2k,x=91-3k,\;\;k \in \mathbb{Z}$ $xy>x+y \; \Leftrightarrow (x-1)(y-1)>1$ $(90-3k)(2k)>1 \; \Rightarrow (30-k)k>0$ $1\le k\le 29$