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.

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 $a,b,c,d,e$.
