I find $(x,y,z) \in \{(1,0,2),(1,0,-2)\}$. I think the solution must to use $\mod 9$.

shinichiman wrote: I find $(x,y,z) \in \{(1,0,2),(1,0,-2)\}$. I think the solution must to use $\mod 9$. Not really. If y is odd, then $11^y\equiv{2}(mod3),$ so $3^x+11^y\equiv{2}(mod3)$( suppose $x\ge1$), impossible. If $x=0$, then the equation becomes $11^y+1=z^2$, which is easy since it can be rewritten as $(z-1)(z+1)=11^y$. So, y is even, $y=2a$. The relation becomes $3^x=(z-11^a)(z+11^a)$, and we have $z-11^a=3^m, z+11^a=3^n, m+n=x$, $11^a=\frac{3^n-3^m}{2}$, so we need $m=0$, which is equivalent to $z=11^a+1, 2.11^a+1=3^x$, now we look at $2.11^a+1=3^x$$(1)$, with solution $a=2,x=5$, which gives $z=122, a=4, x=5$ and $x=1, a=0$ which gives $x=1, y=0, z=2$ We will prove that $(1)$ doesn't have other solutions. We easy get $mod 11$ that $x\equiv{0}(mod5)$, so $x=5t$(i supposed $y\ge 3$, and it becomes $2.11^a=(3^t-1)(3^{4t}+3^{3t}+3^{2t}+3^t+1)$, and we get $3^t-1=2.11^u, 3^{4t}+3^{3t}+3^{2t}+3^t+1=11^v, u+v=a$ If $u=0$, we get $t=1$, already solution, so $u\ge 1$, so $3^t\equiv{1}(mod11)$, so $3^{4t}+3^{3t}+3^{2t}+2^t+1=\equiv{5}(mod11)$, impossible.

First look at $(mod 3)$, we can see $y$ is even. Then $y=2a$. The equation becomes $$3^x=(z-11^a)(z+11^a)$$Then $z-11^a=3^m$ and $z+11^a=3^n$. Then $2\cdot11^a=3^m-3^n$. Then $n=0$. We can easily see that $2\cdot11^2=3^5-1$. By Zsigmondy Theorem, there is always a primitive prime $p$ such that $p|3^m-1^m$ but doesn’t divide $2$ or $11$ for $m\geq5$. So there is no solution for $m\geq5$. So the only solution is $$(x,y,z)=(5,4,122)$$$\blacksquare$