Happy new year 2021

Given that:

$\begin{aligned} x + y + yx & = 2021 \\ xy + x + y + 1 & = 2022 \\ (x+1)(y+1) & = 2022 = 2 \times 3 \times 337 \end{aligned}$

Note that $2$ , $3$ , and $337$ are primes. Therefore $2022 = 2^\red 1 \times 3^\red 1 \times 337^\red 1$ has $(\red 1 +1) (\red 1 +1) (\red 1 +1) = 8$ divisors of $x+1$ and $y+1$ . Hence there are $\boxed 8$ ordered-pair $(x,y)$ solutions.

$\Large \purple {Happy \ New \ Year}$