Rewrite the condition: \[ 3^x \cdot 7^x=(z-2^y)(z+2^y) \] Since $ z-2^y, z+2^y $ can't be the multiples of 3,7 at the some time, so we have two cases. Case: $ \boxed{z-2^y=1,z+2^y=21^x} $ We have $ 2^{y+1}=21^x-1 $. Notice that $ 5|(21^x-1) $, thus, there isn't any solution. Case: $ \boxed{z-2^y=3^x,z+2^y=7^x} $ In this case, we have $ 2^{y+1}=7^x-3^x $, which can be solve immediately by Zsigmody. The only solution is $ x=1,y=1 \rightarrow \{x,y,z\}=\{1,1,5\}$

BSJL wrote: In this case, we have $ 2^{y+1}=7^x-3^x $, which can be solve immediately by Zsigmody. The only solution is $ x=1,y=1 \rightarrow \{x,y,z\}=\{1,1,5\}$ By Zsigmondy, more generally $a^n+b^n$ has at least two prime divisors when $a>b>0, n\ge 2, (a,b,n)\neq (2,1,3)$. $a^n - b^n$ has at least two prime divisors when $a>b+1>1, n\ge 2, (a,b)=1, (a,b,n)\neq (a, 2^m-a, 2)$ for any integer $m$. Here $7>3+1>1, (7,3)=1, (a,b,n)\neq (a,2^m-a,2)$ for any integer $m$, since $x\neq 2$ by checking. $x\ge 2$ gives contradiction by above Zsigmondy lemma, so $x\in\{0,1\}$. $x\neq 0$. $x=1$ gives $(x,y,z)=(1,1,5)$. Don't confuse set notation. It is false that $\{x,y,z\}=\{1,1,5\}$ is a solution (clearly $z\neq 1$). Zsigmondy is overkill here (the proof of Zs isn't elementary). $x$ even gives $2^{y+1}=(7^{x/2}-3^{x/2})(7^{x/2}+3^{x/2})$. $7^{x/2}+3^{x/2}\equiv 2\pmod {4}$, so $7^{x/2}+3^{x/2}=2$ and $x=0$, contr. So $x$ is odd. mod $8$ gives $7^x-3^x\equiv 4\pmod {8}$, so $y=1$, and $(x,y,z)=(1,1,5)$.

ComplexPhi wrote: Solve in nonnegative integers the following equation : $$21^x+4^y=z^2$$ If $x = 0$ we get $(z - 1)(z + 1) = 4^y$,if we look at equation $modulo$ $4$ we obtain,if $z-1 \equiv 0($ $mod$ $4)$ $\implies$ $z+1 \equiv 0$ $(mod$ $4)$,and similarly if $z - 1 \equiv 0$ $(mod$ $4)$ we obtain contradiction so $x \neq 0$. If $y = 0$ we get $(z - 1)(z + 1) = 21^x$,$z-1\neq 1$ and $z+1\neq 1$. If both $z-1$,$z+1$ are divisible with 21 we obtain that $gcd(z-1,z+1)\leq 2$,contradition. If $z - 1 = 3^x$,$z+1 = 7^x$ $\implies$ $7^x - 3^x = 2$,contradiction $7^x - 3^x$ is divisible with 4. If $z - 1 = 7^x$,$z+1 = 3^x$ $\implies$ $7^x < 3^x$,contradiction. We obtain that $y\neq 0$. $3^x\cdot7^x = (z - 2^y)(z+2^y)$. If $z - 2^y = 1$ $\implies$ $z = 2^y + 1$ and $4^y+2^{y+1}+1=4^y+21^x$ or $21^x-1^x = 2^{y+1}$,but $21^x - 1^x$ if divisible with $20$,contradiction. If $z +2^y = 1$ $\implies$ $y = 0$,contradiction. If both $z - 2^y$ and $z + 2^y$ are divisible with 21 $\implies$ $21\equiv 0$ $(mod$ $2)$,contradiction. If $z - 2^y = 3^x$ and $z + 2^y = 7^x$ $\implies$ $3^x + 2^{y+1} = 7^x$. If $y\ge 2$ $\implies$ $3^x\equiv 0$ $(mod$ $8)$ but $3^x\equiv 3,1$ $(mod$ $8)$ and $7^x\equiv -1,1$ $(mod$ $8)$,contradiction $\implies$ $y = 1$,$x = 1$ and $z = 5$. If $z - 2^y = 7^x$ and $z + 2^y = 3^x$ $\implies$ $3^x > 7^x$,contradiction. So the single solution is $(x,y,z) = (1,1,5)$.

To fill in, here is an elementary way to solve $2^{y+1} = 7^x - 3^x$ for $y\geq 2$. By mod $8$ we get that $x$ is even and so $2^{y+1} = (7^{x/2} - 3^{x/2})(7^{x/2} + 3^{x/2})$, thus (as the two factors have gcd $2$) $7^{x/2} - 3^{x/2} = 2$, which is impossible since $3^{x/2} \geq 9$ and $(7/3)^{x/2} - 1 > 5$ for $x\geq 4$.