$\textbf{A) }$ Working in modulo $10$ , we obtain:
The last digit of $2020^x$ is $0$ for all $x\in\mathbb{N}$ .

The last digit of $2^y$ is $\begin{cases}2,\text{ for y }\equiv 1\pmod4\\4,\text{ for y }\equiv 2\pmod4\\8,\text{ for y }\equiv 3\pmod4\\6,\text{ for y }\equiv 0\pmod4\end{cases}$

The last digit of $2024^z$ is $\begin{cases}4,\text{ for z }\equiv 1\pmod2\\6,\text{ for z }\equiv 0\pmod2\end{cases}$

Results: $y=2w,\;w\in\mathbb{N}$ and $w-z\equiv0\pmod2$ .
The equation becomes:
$2020^x+4^w=2024^z$ , where $w,z$ have the same parity $\quad\textbf{(1)}$ .

$\textbf{B) }$ Working in modulo $3$ , we obtain:
$2020^x\equiv1\pmod3;\;4^w\equiv1\pmod3\Longrightarrow 2024^z\equiv2\pmod3\Longrightarrow z$ is odd number $\Longrightarrow w$ is odd number.

$\textbf{Case 1: }\min\{x,w,z\}=w;\;w\le x;\;w\le z$ .
Dividing in $\textbf{(1)}$ by $4^w$ results:
$505^x\cdot 4^{x-w}+1=506^z\cdot 4^{z-w}$ .
$z-w$ is an even non-negative number and results: the last digit of $506^z\cdot 4^{z-w}$ is $6$ .
The last digit of $505^x\cdot 4^{x-w}$ must be $5$ , hence $x=w$ and we obtain
$505^x+1=506^z\cdot 4^{z-w}$ .
Working in modulo $4$ in the last relation, results:
$505^x\equiv1\pmod4\Longrightarrow 506^z\cdot 4^{z-w}\equiv2\pmod4\Longrightarrow$
$\Longrightarrow z-w=0;\;z=1\Longrightarrow x=w=z=1\Longrightarrow y=2$
and the triplet $(x,y,z)=(1,2,1)$ is solution of the equation $2020^x+2^y=2024^z$ .

$\textbf{Case 2: }\min\{x,w,z\}=x;\;x<w;\;x\le z$ .
Dividing in $\textbf{(1)}$ by $4^x$ results:
$505^x+4^{w-x}=506^z\cdot 4^{z-x}$ .
$505^x$ is odd number; $4^{w-x}$ and $506^z\cdot 4^{z-x}$ are even numbers, hence the last equation has no solutions.

$\textbf{Case 3: }\min\{x,w,z\}=z;\;z<x;\;z<w$ .
$2020^x+4^w=2024^z$ .
$v_2(2024^z)=3z$ , odd number (see $\textbf{B)}$ ).

$\textbf{case 3.1: }x<w$
$2020^x+4^w=4^x(505^x+4^{w-x})\Longrightarrow v_2(2020^x+4^w)=2x$ , even number, hence the equation has no solutions.

$\textbf{case 3.2: }x>w$
$2020^x+4^w=4^w(505^x\cdot4^{x-w}+1)\Longrightarrow v_2(2020^x+4^w)=2w$ , even number, hence the equation has no solutions.

$\textbf{case 3.3: }x=w$
$2020^x+4^x=2024^z$ .
$z=2u+1;\;x=w>z$ , where $u\in\mathbb{N}\cup\{0\}$ (see $\textbf{B)}$ ).
$2020^x+4^x=4^x(505^x+1)$ .
$505^x+1\equiv2\pmod4\Longrightarrow v_2(505^x+1)=1\Longrightarrow$
$\Longrightarrow v_2(2020^x+4^x)=v_2(4^x(505^x+1))=2x+1=3z\Longrightarrow$
$\Longrightarrow 2x+1=6u+3\Longrightarrow x=w=3u+1;\;z=2u+1$ .
$x>z\Longrightarrow u>0$ .
The equation becomes:
$2020^{3u+1}+4^{3u+1}=2024^{2u+1}\Longrightarrow 505^{3u+1}+1=2\cdot253^{2u+1}\Longrightarrow$
$\Longrightarrow 505\cdot505^{3u}+1=506\cdot253^{2u}$ , contradiction since
$505\cdot505^{3u}>506\cdot253^{2u},\;\forall u\in\mathbb{N}$ (proved many times in the previous posts).

$\textbf{Conclusion:}$
The equation in positive integers $2020^x+2^y=2024^z$ has the unique solution $(x,y,z)=(1,2,1)$ .