$\rm 3\not {|} 2^x+21^y=z^2\Rightarrow z^2\equiv 1\pmod 3$.So $\rm (-1)^x\equiv 1\pmod 3\Leftrightarrow 2\mid x$.Let $\rm x=2p,p\in \mathbb{N^*}$. $\rm 2^{2p}+21^y=z^2\Leftrightarrow 21^y=(z-2^p)(z+2^p)$.Let $\rm d=(z-2^p,z+2^p)$ $\rm \left\{\begin{matrix} &\rm d\mid z-2^p & \\ &\rm d\mid z+2^p & \end{matrix}\right.\Rightarrow d\mid 2^{p+1}$.But also $\rm d\mid z-2^p\mid 21^y$ so $\rm d=1$. We have 4 cases: Case 1: $\rm \left\{\begin{matrix} &\rm z-2^p=21^y & \\ & \rm z+2^p=1 & \end{matrix}\right.\Rightarrow 2^{p+1}=1-21^y<0$,contradiction Case 2: $\rm \left\{\begin{matrix} & \rm z-2^p=7^y & \\ & \rm z+2^p=3^y & \end{matrix}\right.\Rightarrow 2^{p+1}=3^y-7^y<0$.contradiction. Case 3: $\rm \left\{\begin{matrix} & \rm z-2^p=3^y& \\ & \rm z+2^p=7^y& \end{matrix}\right.\Rightarrow 2^{p+1}=7^y-3^y$. If $\rm 2\not {|} y$ then $\rm 7^y-3^y\equiv -1-3\equiv 4\pmod 8$ so $\rm p=1$. $\rm 4=7^y-3^y\Leftrightarrow y=1$,because $\rm 7^y-3^y>6^y-3^y=3^y(2^y-1)>3^y$. So $\rm y=1,x=2p=2,z=3^1+2^1=5$. If $\rm 2\mid y$ ,let $\rm y=2r,r\in \mathbb{N^*}$.We have that $\rm 2^{p+1}=(7^r-3^r ) (7^r+3^r)$. Let $\rm m=(7^r-3^r,7^r+3^r)$.Then $\rm m\mid 2\cdot 3^r$,but also $\rm m\mid 2^{p+1}$ so since $\rm 2\mid m$ we have $\rm m=2$. So $\rm 7^r+3^r=2$,contradiction. Case 4: $\rm \left\{\begin{matrix} & \rm z-2^p=1& \\ & \rm z+2^p=21^y& \end{matrix}\right.\Rightarrow 2^{p+1}=21^y-1$.We have $\rm 2^{p+1}\equiv -1\pmod 7$ but $\rm ord_7(2)=3$ so $\rm 2^t\equiv 1,2,4$,contradiction. So the only solution is $\rm (x,y,z)=(2,1,5)$

You have a small mistake ans as a result you lose a solution. If y=0 we have: 2^x=(z-1)(z+1) Because (z-1, z+1)=2 we have z-1=2. Or. z-1=2^(x-1) z+1=2^(x-1). z+1=2 The first give the solution (x,y,z):(3,0,3) and the second does not have a solution If y bigger than zero we can take mod 3.......... So all the solution is (x,y, z):(3,0,3),(2,1,5) Also you can prove that 2^(p+1)=7^y-3^y more easily if you get the zigmondy theory

P2nisic wrote: You have a small mistake ans as a result you lose a solution. If y=0 we have: 2^x=(z-1)(z+1) Because (z-1, z+1)=2 we have z-1=2. Or. z-1=2^(x-1) z+1=2^(x-1). z+1=2 The first give the solution (x,y,z):(3,0,3) and the second does not have a solution If y bigger than zero we can take mod 3.......... So all the solution is (x,y, z):(3,0,3),(2,1,5) Also you can prove that 2^(p+1)=7^y-3^y more easily if you get the zigmondy theory it said that $x,y,z$ positive integers, so $y\neq 0$

Yes . sorry i haven't see that

Took me 1 hr to solve lol. Cute!