1. The equation is written:
$$xy(x+y-10)-3x^2-2y^2+21x+16y=60 \\ \Leftrightarrow (y-3)x^2+(y^2-10y+21)x=2y^2-16y+60 \\ \Leftrightarrow (y-3)x^2+(y-7)(y-3)x=2(y-3)(y-5)+30 \\ \Leftrightarrow (y-3)x^2+(y-7)(y-3)x-2(y-3)(y-5)=30 \\ \Leftrightarrow (y-3)[x^2+(y-7)x-2(y-5)]=30 \\ \Leftrightarrow (y-3)(x^2-7x+xy-2y+10)=30 \\ \Leftrightarrow (y-3)(x^2-4-7x+xy-2y+14)=30 \\ \Leftrightarrow (y-3)((x-2)(x+2)-7(x-2)+(x-2)y)=30 \\ \Leftrightarrow (y-3)(x-2)(x+y-5)=30=1 \cdot 2 \cdot 3 \cdot 5.$$
Since $x-2$, $y-3$ and $x+y-5$ are integers such that $x-2+y-3=x+y-5$
and moreover $x-2 \ge -1$, $y-3 \ge -2$ and $x+y-5 \ge -3$, we have to consider the following cases:
$$\begin{cases} x-2=2 \\ y-3=3 \\ x+y-5=5 \end{cases} \text{ or } \begin{cases} x-2=3 \\ y-3=2 \\ x+y-5=5 \end{cases} \text{ or } \begin{cases} x-2=1 \\ y-3=5 \\ x+y-5=6 \end{cases} \text{ or } \begin{cases} x-2=5 \\ y-3=1 \\ x+y-5=6 \end{cases} \\ (x,y) = (4,6) \text{ or } (x,y) = (5,5) \text{ or } (x,y) = (3,8) \text{ or } (x,y) = (7,4).$$

Alternatively, from the equation $(y-3)x^2+(y-3)(y-7)x=2(y^2-8y+30)$ we get:
$$(y-3)[x^2+(y-7)x]=2y^2-16y+60 \quad (1)$$
From which, since $x$, $y$ are positive integers, arises that: $(y-3)|2y^2-16y+60$. Since $2y^2-16y+60=(y-3)(2y-10)+30$ arises that $y-3|30$. Therefore $y-3 \in \{\pm1, \pm2, \pm3, \pm5, \pm6, \pm10, \pm15, \pm30\}$, and since $y-3 \ge -2$ we get: $y-3 \in \{\pm1, \pm2, 3, 5, 6, 10, 15, 30\}$.
For each of these values, by substituting to (1) we obtain a trinomial of $x$.