Let $A=20^x$, $B=14^{2y}$, $C=(x+2y+z)^{zt}$. Since the LHS is even, we have $2|x+2y+z$ hence $x$ and $z$ have the same parity. If $x$ and $z$ are both even, then $A$ and $B$ are square numbers not divisible by $3$, so $A+B\equiv 2\pmod 3$, whereas $C$ is also a perfect square, contradiction. So $x$ and $z$ are odd, and $3|x+2y+z$. Now look at the equation modulo $7$. Since $x$ is odd, we have $A\equiv (-1)^x\equiv -1\pmod 7$, so $C=A+B\equiv -1\pmod 7$, where $-1$ is a non-quadratic residue modulo $7$, hence $zt$ must be odd. Suppose $zt=1$, then we have $A+B=x+2y+1$. But by Bernoulli's inequality, $A\ge 1+19x$ and $B\ge 1+26y$; adding these two gives $A+B>x+2y+1$, a contradiction. Hence $zt\ge 2$, so if $3|x+2y+z$, then we have $9|C$. Also, $A\equiv (-7)^x\equiv -7^x\pmod 9$ and $B=196^y\equiv 7^y\pmod 9$, so we get $7^x\equiv 7^y\pmod 9$. Since the order of $7$ modulo $9$ is $3$, this implies that $x\equiv y\pmod 3$, and so $3|x+2y$, yielding $3|z$. Note that if $x\equiv y\equiv 0\pmod 3$, then $A,B,C$ are perfect cubes and the equation cannot hold due to Fermat's Last Theorem (or at least the $n=3$ case proven by Euler). Suppose $x\neq y$, say $x<y$. Then $A+B=4^x\cdot 5^x+4^y\cdot 49^y=4^x\left(5^x+4^{y-x}\cdot 49^y\right)$, hence $v_2(A+B)=2x$. Therefore $zt|2x$. Since $3|z$, we get $3|x$. But this has just been proven impossible. We get a similar absurdity if $y<x$. Therefore $x=y$ and the equation becomes $20^x+196^x=(3x+z)^{zt}$. Using LTE, we get $v_3(20^x+196^x)=v_3(216)+v_3(x)=3$ because $x$ is not divisible by $3$. Hence our equation may only hold if $zt|3$. But since $3|z$, only $z=3$, $t=1$ is possible. What remains is $20^x+196^x=(3x+3)^3$. Note that $x=1$ is a solution, suppose there is a solution $x\ge 2$. Then by Bernoulli, we get $196^x>125^x=(5^x)^3\ge (1+4x)^3\ge (3+3x)^3$. Therefore the only solution is $(x,y,z,t)=(1,1,3,1)$.