We have the following cases: $1) y>2x$. Then we get that: $2^{2x}(505^x+2^{y-2x})=2^{3z}253^z$,so $2x=v_2(LHS)=v_2(RHS)=3z$ and hence $505^x+2^{y-2x}=253^z$, which is a contradiction by $mod 3$ $2) y<2x$. Then we get that: $2^y(2^{2x-y}505^x+1)= 2^{3z}505^z$ With the same way as case $1$ we get that $y=3z$ and hence $2^{2x-y}505^x+1=253^z$, which is a contradiction by $ mod 3$. $3) y=2x$. Then we get that: $2^{2x}(505^x+1)=2^{3z}253^z$ $505^x+1 \equiv 2 mod 4$, and hence $2x+1=v_2(LHS)=v_2(RHS)=3z$, so equivalently we have that $505^x+1=2*253^z=2*253^{\frac {2x+1} {3}}$, which has only $x=1$ as a positive integer solution. So, in this case $(x,y,z)=(1,2,1)$ ,which is the only solution of the given equation.

Consider $U_2$ we get: $2x=3z$ contradiction since then $LHS>RHS$ $y=3z$ then $2020^x=2024^z-8^z=2016[...]$ contradiction since $7|2016$ but not $2020$ $y=2x$ we have that $2^{2x}(505^x+1)=2^{3z}*253^z$ or $505^x+1=2^{3z-2x}*253^z$ Consider $mod4$ we get that $3z-2x=1$ and then from inelyalites esily get $x=1,y=2,z=1$

First,assume that $x \geq 2 $ and $y \geq 3$. In first case we will inspect $2x=y$. Then,from taking both sides' power, $2x=y=3z$. It is known that,$x>z$ but, $505^x+1=253^z$ which is contradiction. In second case we will inspect $2x>y$. Like first method,we get $y=3z$ then, $2x>3z$. So, $2^{2x-y}.505^x+1 \geq 2.505^x+1 >253^z$ contradiction again. In third case we will look $y>2x$. From getting power both sides, we found that,$2x=3z$.$505^x+2^{y-2x}=253^z$ obviously contradicition. So,$x=1,y=2,z=1$.

Proposed by Serbia. Do not get negatively fooled by the classical looking statement -- the various possible approaches here can teach students a lot of important ideas!

Proposed by me (Ognjen Tešić, Serbia).

Taking the equation modulo $5$ yields $y$ is even, so $y=2y_1$. Now if $x\neq y_1$ we have \[2x\geq \nu_{2}(2020^x+2^y)=\min\{2x, 2y_1\}=\nu_{2}(2024^z)=3z.\]However, $2x<3z$ as $2024^z>2020^x>2024^{\frac{2}{3}x}$. Therefore $x=y_1$, and the equation becomes $4^x(505^x+1)=2024^z$. Modulo $11$ implies $x$ is odd, at which point $\nu_2(505^x+1)=\nu_2(506)=1$. Comparing the $\nu_2$'s of both sides gives $2x+1=3z$, so $x=3t+1$, $z=2t+1$ for some nonnegative integer $t$. Clearly, $t=0$ leads to the solution $(x, y, z) = (1, 2, 1)$. When $t>0$ we derive a contradiction from: \[1>\frac{2020^{3t+1}}{2024^{2t+1}}=\frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)^t\geq \frac{505}{506}\cdot \left(\frac{2020^3}{2024^2}\right)>1.\]

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The only solution is $(x,y,z) = (1,2,1)$, which obviously works. We now prove it is the only one. Claim: $y=2x$. Proof: Take both sides of the equation modulo $3$ to get $1+2^y\equiv 2^z \pmod{3}$. Clearly, we must have $2^y \equiv 1 \pmod{3}$, so $y$ is even and $z$ is odd. Let $y = 2y_0$, for a positive integer $y_0$. Then, the given equation becomes $2020^x + 4^{y_0} = 2024^z$. Note that $v_2(2024^z) = 3z$ is always odd. Since $v_2(4)$ and $v_2(2020)$ are both even, it follows if $v_2(2020^x)\ne v_2(4^{y_0})$, then either $v_2(2020^x + 4^{y_0}) = v_2(2020^x) = 2x$ or $v_2(2020^x + 4^{y_0}) = v_2(4^{y_0})=2y_0$, neither of which are odd. Thus, $v_2(2020^x) = v_2(4^{y_0})$. Because $v_2(2020)=v_2(4)$, we must have $x=y_0$, and our claim follows. Claim: The only possible value of $x$ is $1$. Proof: Rewrite the given equation as $4^x(505^x + 1^x) = 2024^z$. Note that when $x=1$, $505^1+1^1=506$ is divisible by all the prime factors of $2024$, which are $2$, $11$, and $23$. If $x>1$, by Zsigmondy, there exists some $p\notin \{2,11,23\}$ that divides $505^x + 1^x$, and hence it is impossible that $505^x+1^x \mid 2024^z$. Thus, $x=1$, and we are done. $\blacksquare$

The only solution is $(1,2,1)$, which clearly works. Notice that taking the equation mod $7$ gives $4^x + 2^y \equiv 1 \pmod 7$. Since $4^3 \equiv 2^3 \equiv 1\pmod 7$, if $3$ divided either one of $x$ or $y$, then we have that one of $4^x, 2^y$ is $0\pmod 7$, which is absurd. Hence $3\nmid xy$. Hence $\nu_2(2020^x), \nu_2(2^y)$ are both not multiples of $3$, so they cannot be equal to $ \nu_2(2024^z)$. If $\nu_2(2020^x) \ne \nu_2(2^y)$, then we would have $\nu_2(2024^z) = \nu_2(2020^x + 2^y) \in \{\nu_2(2020^x) , \nu_2(2^y)\}$, which is absurd, so $\nu_2(2020^x) = \nu_2(2^y)$, so $2x = y$. Now we have \[ 2020^x + 4^x = 2024^z \]If $x > 1$, then by Zsigmondy there exists a prime $p$ not dividing $2020^1 + 4^1 = 2024$ that divides $2020^x + 4^x$, absurd. Hence $x = 1$ must hold.

$\textbf{A) }$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 $(x,y,z)=(1,2,1)$.

take mod 3 to get $1+(-1)^y=(-1)^z$. Because of this, we know y must be even and z must be odd. Let $y=2y_1$ and $z=2z_1+1$. Prime factorizing the original equation, $2^{2x}\cdot505^x+2^{2y_1}=2^{6z_1+3} \cdot253^{2z_1+1}$ Now considering the largest factor of 2 that divides the LHS, if $2y_1$ is the smallest factor, then $2y_1=6z_1+3$, impossible. Similarly, if 2x is the largest factor of 2, $2x=6z_1+3$ is impossible. This means $2x=2y_1$ and the factor of 2 on the LHS is $2^{2x+1}$ since the v2 of $505^{x}+1$ is at 2. (since its 0 mod 2 and not 0 mod 4). So we have $2x+1=3z_1$ Going back, we have $x=x, y=2x, z=\frac{2x+1}{3}$. since $\frac{2x+1}{3}$ is an integer, set $x=3k+1$ Plugging in and dividing by largest factor of 2 we get $505^{3k+1}+1=2\cdot253^{2k+1}.$ k=0 works, but any larger k dosent work since $505^{3k+1}+1 > 505 \cdot 505^{3k} > 506 \cdot 253^{2k}$ so the only solution is k=0, which represents $(x,y,z)=(1,2,1)$