Here is another approach.

(1,2) and (2,1) are pythagorean triples and are parametrized.

Set $x_1=x,x_{p+1}=x-1$. Then $x_1^2-x_{p+1}^2=2x-1$.

$2x-1$ represents all positive odd integers, so it is difference of two squares and sum of two squares infinitely often.

If $p>1$, set all variables other than $x,x_2,x_{2+p}$ to $4$.

For integer $m$, we have $2x-1+4m$ difference of two squares and this has infinitely many solutions.

If $p=1$, set all variables other than $x,x_{2+p},x_{3+p}$ to $4$.

For integer $m$, we have $2x-1+4m$ sum of two squares and this has infinitely many solutions, since it is prime of the form $4k+1$ infinitely often by primes in arithmetic progressions.

Here is another approach.

(1,2) and (2,1) are pythagorean triples and are parametrized.

Set $x_1=x,x_{p+1}=x-1$. Then $x_1^2-x_{p+1}^2=2x-1$.

$2x-1$ represents all positive odd integers, so it is difference of two squares and sum of two squares infinitely often.

If $p>1$, set all variables other than $x,x_2,x_{2+p}$ to $4$.

For integer $m$, we have $2x-1+4m$ difference of two squares and this has infinitely many solutions.

$$ 2x-1+4m + x_2^2-x_{2+p}^2=0$$

If $p=1$, set all variables other than $x,x_{2+p},x_{3+p}$ to $4$.

For integer $m$, we have $2x-1+4m$ sum of two squares and this has infinitely many solutions, since it is prime of the form $4k+1$ infinitely often by primes in arithmetic progressions.

$$2x-1+4m-x_{2+p}^2-x_{3+p}^2=0$$

To avoid factorization in sum of two squares, find prime in the arithmetic progression.