Is there any solution for $n^x+(n+d)^y+(n+2d)^z=a^b$ 
Is there any solution for $n,d,a,b,x,y,z,b\in \mathbb{Z}_+$ with $x,y,z,b\ge 3$ and $\{n,n+d,n+2d,a\}$ has least one common prime factor s.t.
$$n^x+(n+d)^y+(n+2d)^z=a^b$$

I think this problem help to enlarge Beal conjecture.

Related post
Can it be shown, $n^4+(n+d)^4+(n+2d)^4\ne z^4$?
Extending Fermat's Last Theorem
 A: EDIT : answer with $d>0$. 
Yes, there is !
Let $d=a=n=36$ and $x=y=z=3$ and $b=4$.
A: Here is a fairly general solution, including for where you have more terms on the left side. Consider your Normalising Beal's conjecture question's equation of
$$\sum_{q=0}^{u}(n+qd)^{m_{q}} = a^b \tag{1}\label{eq1A}$$
Choose any $u \ge 1$ and $c \gt 2$. Then have 
$$m_{q} = c, \; 0 \le q \le u \tag{2}\label{eq2A}$$
Next, have
$$n = d = \sum_{q=0}^{u}(1+q)^c  \tag{3}\label{eq3A}$$
The LHS of \eqref{eq1A} now becomes
$$\begin{equation}\begin{aligned}
\sum_{q=0}^{u}(n+qd)^{m_{q}} & = \sum_{q=0}^{u}(n+qd)^{c} \\
& = \sum_{q=0}^{u}n^c(1+q)^{c} \\
& = n^c\left(\sum_{q=0}^{u}(1+q)^{c}\right) \\
& = n^c(n) \\
& = n^{c+1}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, you can choose $a = n$ and $b = c + 1$ so this matches the RHS of \eqref{eq1A}.
Your question here uses $u = 2$ in \eqref{eq1A}. Also, this answer is a special case of my solution, with it using $c = 3$. Thus, based on \eqref{eq3A}, you get $n = d = 1 + 2^3 + 3^3 = 1 + 8 + 27 = 36$, with the rest following as stated in that answer.
