<s>[INST] Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n=2$,
$$
x^2+y^3=z^4
$$
I'm already pretty lost.
After this it gets even more complicated, has anyone encountered this problem before?
I want all the solutions, or at least infinitely many for every equation.
 [/INST] If $n+2$ is composite, say $n+2=mk$ with $m,k>1$, take any $z$, $x_m = z^k$, all other $x_j = 0$. 
EDIT:
Suppose we have a solution $(a_1, \ldots, a_{n+1})$ in positive integers with 
$a_1^{d_1} + \ldots + a_n^{d_n} = a_{n+1}^{d_{n+1}}$, and positive integer $k$ so that $d_i+k$ are pairwise coprime.  By the Chinese Remainder Theorem there are positive integers $m_1, \ldots, m_{n+1}$  such that $m_i \equiv 0 \mod d_j + k$ for $j \ne i$ while $m_i \equiv k \mod d_i + k$.  Let $b = a_1^{m_1}\ldots a_{n+1}^{m_{n+1}}$ and $t_i = (b/a_i^k)^{1/(d_i+k)}$, which is a positive integer. 
Then 
$$(t_1 a_1)^{d_1+k} + (t_2 a_2)^{d_2+k} + \ldots + (t_n a_n)^{d_n+k}
= b (a_1^{d_1} + \ldots + a_n^{d_n}) = b a_{n+1}^{d_{n+1}} = (t_{n+1} a_{n+1})^{d_{n+1}+k}$$
For example, starting from 
$$ 28^2 + 8^3 = 6^4$$ ($a_1 = 28$, $a_2 = 8$, $a_3 = 6$), for $k=1$ 
where $3,4,5$ are pairwise coprime (it will work for any odd $k$) we take $m_1,m_2,m_3 = 40, 45, 36$.
Then $$\eqalign{b &= 28^{2} 8^3 6^4 = 2^{251} 3^{36} 7^{40} \cr
    t_1 &= (b/a_1)^{1/3} = 2^{83} 3^{12} 7^{13}\cr
    t_2 &= (b/a_2)^{1/4} = 2^{62} 3^9 7^{10}\cr
    t_3 &= (b/a_3)^{1/5} = 2^{50} 3^7 7^8\cr
    (t_1 a_1)^{3} &+ (t_2 a_2)^4 = (t_3 a_3)^5 \cr
    0^2 + (2^{85} 3^{12} 7^{14})^3 &+ (2^{65} 3^9 7^{10})^4 = (2^{51} 3^8 7^8)^5 \cr}$$
</s>