what is written below is a conjecture that I posed , and I ask for a proof or a disproof of it .I have checked  the conjecture from $n$=$1$ up to $n$=$10$  using Matlab, and all results were in agreement with the conjecture .
The conjecture is as follows : 
assume $x$ is a positive real parameter that does not equal $1$ , and assume $y$ and $z$ are non-zero real parameters , and consider for all $i, j \in \mathbb N$,
$$a(i,j) = \frac{(x^{yi+z} + 1)^{j-1} + (x^y-1)}{x^y}$$ ; then for all $n \in \mathbb N$ , the solution set of the matrix system 
$[a(i,j) \mid 1 \leq i \leq n, 1 \leq j \leq (1+n)]$ exists and is unique with respect to $n$ and $x$ and $y$ and $z$ ,and each element in it is a sum of powers of $x$ with integer coefficients , and each of these powers of $x$ has the power as a linear combination of $y$ and $z$ such that the coefficients of $y$ and $z$ are non-negative integers .

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###Example

For $n$=$4$, the solution set will be :

$s_1=- x^{y+z} - x^{z+2y} - x^{z+3y} - x^{z+4y} - x^{3y+2z} - x^{4y+2z} - 2x^{5y+2z} - x^{6y+2z} - x^{6y+3z} - x^{7y+2z}$ $\qquad- x^{7y+3z} - x^{8y+3z} - x^{9y+3z} - x^{9y+4z} - 1$


$s2$=3$x^{y+z}$ + 3$x^{z+2y}$ + 3$x^{z+3y}$ + 3$x^{z+4y}$ + 2$x^{3y+2z}$ + 

$\qquad 2x^{4y+2z}$ + 4$x^{5y+2z}$ + 2$x^{6y+2z}$ + $x^{6y+3z}$ + 2$x^{7y+2z}$ + $x^{7y+3z}$ + $x^{8y+3z}$ + $x^{9y+3z}$ + 4


$s3$=- 3$x^{y+z}$ - 3$x^{z+2y}$ - 3$x^{z+3y}$ - 3$x^{z+4y}$ - $x^{3y+2z}$ - $x^{4y+2z}$ - 2$x^{5y+2z}$ - $x^{6y+2z}$ - $x^{7y+2z}$ - 6


$s4$=$x^{y+z}$ + $x^{z+2y}$ + $x^{z+3y}$ + $x^{z+4y}$ + 4

Another example that explains the previous example by assuming $x$=$2$,$y$=$1$,and $z$=$1$:
we will have for all $i, j \in \mathbb N$,
$a(i,j)$ = ($(2^{i+1} + 1)^{j-1}$ + 1)/2

since $n$=$4$,we will find the solution set of the matrix system $[a(i,j) \mid 1 \leq i \leq 4, 1 \leq j \leq 5]$

To solve the matrix system we can represent it as linear system of simultaneous 
equations , the system will be : 

$1$ $s1$ + $3$ $s2$ + $13$ $s3$ + $63$ $s4$ = $313$ ...($1$) ,
$1$ $s1$ + $5$ $s2$ + $41$ $s3$ + $365$ $s4$ = $3281$ ...($2$) ,
$1$ $s1$ + $9$ $s2$ + $145$ $s3$ + $2457$ $s4$ = $41761$ ...($3$) ,
$1$ $s1$ + $17$ $s2$ + $545$ $s3$ + $17969$ $s4$ = $592961$ ...($4$) ;
solving the system we get $s1$=$-17053$ , $s2$=$10104$ , $s3$=$-1306$ , $s4$=$64$ ,so we notice that the values of $s1$,$s2$,$s3$,$s4$ can be found by substituting ($x$=$2$,$y$=$1$,$z$=$1$) in the formulas of $s1$,$s2$,$s3$,$s4$ in the first example .

Thank you .