The New Year Problem.

Solution:

An easy one to see the new year in with.
The first term, ${\left(2019\right)}^0=1$ as any number raised to the zeroth power is one (if it's not zero), and
the second term, ${\left(2020\right)}^1=2020$ as any number raised to the power of one is that number.
So, the first two terms sum to

Now our equation simplifies to $2021{\left(2021\right)}^x=0$,
and further to ${\left(2021\right)}^x=2021$

Hence, $x=1$.

Further Reading:

(care of )