For me, neither Reduce or Solve seem to be able to provide any solutions. However, if we fix x to integer values, FindInstance is able to give us at least a solution. For the first couple of integers we get
{x -> 1, y -> -2018} {x -> 2, y -> -1008} {x -> 3, y -> -2014/3} {x -> 4, y -> -503} {x -> 5, y -> -402} {x -> 6, y -> -1004/3} {x -> 7, y -> -2006/7}
There is a pattern here, and one can check that the following is always solution to your equation
y[x_] := -2020/x + 2
This can be checked analytically
{4^(x + 1) + 2 Log[2, x y + 2021], 2^(x y + 2022) + 2 Log[2, 2 x + 1]} /. y -> y[x] // Simplify (* {4^(1 + x) + (2 Log[1 + 2 x])/Log[2], 4^(1 + x) + (2 Log[1 + 2 x])/Log[2]} *)
so this also holds for real values of x.
For y to be an integer, x needs to be a divisor of 2020, so we get the following set of integer solutions
{{x -> -2020, y -> 3}, {x -> -1010, y -> 4}, {x -> -505, y -> 6}, {x -> -404, y -> 7}, {x -> -202, y -> 12}, {x -> -101, y -> 22}, {x -> -20, y -> 103}, {x -> -10, y -> 204}, {x -> -5, y -> 406}, {x -> -4, y -> 507}, {x -> -2, y -> 1012}, {x -> -1, y -> 2022}, {x -> 1, y -> -2018}, {x -> 2, y -> -1008}, {x -> 4, y -> -503}, {x -> 5, y -> -402}, {x -> 10, y -> -200}, {x -> 20, y -> -99}, {x -> 101, y -> -18}, {x -> 202, y -> -8}, {x -> 404, y -> -3}, {x -> 505, y -> -2}, {x -> 1010, y -> 0}, {x -> 2020, y -> 1}}
However, there may be solutions outside of this set.
Edit
Your equation can be rewritten as \begin{equation} 2^{2x+1}-2^{xy+2021} = \log_2{\left(\frac{2x+1}{xy+2021}\right)} \end{equation}
The solution found above is just the condition for both sides to vanish, i.e.
\begin{equation} 2x+1=xy+2021, \end{equation}
which is a much nicer derivation.