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.
 A: For $n=2$ and $n=3,4$ we can give infinite families using a Pell equation and elliptic curves, respectively,

$n=2$:

$$\big(4q^2(p^2-2)\big)^2+(4dq^2)^3  = (2pq)^4$$
where $p,q$ solve $p^2-d^3q^2=1\tag1$. 

$n=3$:

$$(a y)^2 + (ma)^3 + a^4 = a^5$$
and the elliptic curve solvable for an appropriate constant $m$,
$$a^3 - a^2 - m^3 a = y^2\tag2$$
Example: For $m=2$,
$$\begin{aligned}
(4\cdot4)^2 + (2\cdot4)^3 + 4^4 &=4^5\\
(9\cdot24)^2 + (2\cdot9)^3 + 9^4 &=9^5\\ 
\big(\tfrac{10252a}{125}\big)^2+(2a)^3 + a^4 &= a^5
\end{aligned}$$
and $a=\big(\tfrac{22}{5}\big)^2$. This elliptic curve has an infinite number of rational points and by multiplying by an appropriate factor (such as $5^{60}$), then one can easily clear denominators and get integer solutions. (Note: The initial solution of $(2)$ may have a large height. For example, for $m=13$ we have $a=(\frac{45539}{6612})^2$.)

$n=4$:

Similarly, one can use,
$$(a y)^2 + (2a)^3 + a^4 + a^5 = a^6$$
$$(a y)^2 + (2a)^3 + (3a)^4 + a^5 = a^6$$
both of which entail solving an elliptic curve.

$n>4$: 

A nice identity with almost consecutive exponents is one by Enrico Jabara,
$$(7^3m^{11} n^4)^3 + (7\cdot14^4m^{11}n)^4 + (7m^4n^4)^5 + (m n^4)^7 + (14^4 m^6-n^4)^8 = (14^4 m^6+n^4)^8$$ 
where $m = 28^4$ and arbitrary $n$, and which he probably found by expanding,
$$(an^4)^3 + (bn)^4 + (cn^4)^5 + (dn^4)^7 + (e-n^4)^8 = (e+n^4)^8$$
collecting powers of $n$, and solving for $a,b,c,d,e$.
A: Here's a proof of infinitely many solutions $(x_1, x_2, x_3, \dots, x_n, z)$:
For $n = 1$ we have $(1, 1)$. For $n = 2$ we have $(3^3, 2 \cdot 3^2, 3^2)$.
For $n \ge 3$ we have $(2, 3, 1, 2, 2, 2, 2, \dots, 2)$.
(To verify: $2^2 + 3^3 + 1^4 + 2^5 + 2^6 + \dots + 2^n = 32 + 2^5 (1 + 2 + 4 + \dots + 2^{n-5}) = 32 + 2^5 (2^{n-4} - 1) = 2^{n+1}$, where we have used a geometric series to simplify the sum.)
Thus all $n \ge 1$ yields at least one solution in positive integers. To get infinitely many solutions, multiply term $x_k$ by $m^{\frac{(n+1)!}{k}}$ and $z$ by $m^{n!}$; then both sides of the Diophantine equation are multiplies by $m^{(n+1)!}$. Choosing arbitrary $m \ge 2$ gives infinitely many solutions, as desired.
A: 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 $$\eqalign{b &= 28^{2} 8^3 6^4 = 2^{251} 3^{36} 7^{40} \cr
    t_1 &= (b/a_1)^{1/3} = 2^{83} 3^{12} 7^{13}\cr
    t_2 &= (b/a_2)^{1/4} = 2^{62} 3^9 7^{10}\cr
    t_3 &= (b/a_3)^{1/5} = 2^{50} 3^7 7^8\cr
    (t_1 a_1)^{3} &+ (t_2 a_2)^4 = (t_3 a_3)^5 \cr
    0^2 + (2^{85} 3^{12} 7^{14})^3 &+ (2^{65} 3^9 7^{10})^4 = (2^{51} 3^8 7^8)^5 \cr}$$
A: I also find this problem quite interesting.
An easier question to ask is about the number of solutions for each $n$. Of course, if we find one set $(x_1, x_2, ... ,x_n)$, then for any integer $k$ we set $p_i = k^{\frac{\prod_{t = 1}^{n+1} t }{i}}$, then
$$(p_1x_1,p_2x_2, ... ,p_nx_n)$$ is also a solution, thus wielding an infinte number of them.
So the question is whether or not there exists at least one solution for each n.
Another idea would be to work inductively. Suppose $x_1^2 + x_2^3 + ... + x_{n-1}^n = z^{n+1}.$ Then $x_1^2 + x_2^3 + ... + x_{n-1}^n + x_n^{n+1} = z^{n+1} + x_n^{n+1}$; Thus the question becomes whether or not we can find two integers $a,b$ such that given $z$ the relation $z^{n+1} + a^{n+1} = b^{n+2}$ holds. 
Does anybody have any idea how can one prove an infinite number of solutions for any $n$?
