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The answer is yes and appears in R. Cohn's book 'Difference algebra' Lemma II p. 201.

Try for another proof by induction on the complexity of the differential equation:

Let $(R,\sigma)$ be a difference domain (i.e. a domain equiped with a ring morphism $\sigma:R\rightarrow R$ satisfying $\sigma(a+b)=\sigma a+\sigma b$ and $\sigma(ab)=(\sigma a)(\sigma b$). Let $\bar x=(x_1,\dots,x_n)$. Let $R\{\bar x\}$ be the set of difference polynomials in variables $\bar x$, that is, the set of polynomials in variables $\{\sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\}$, $$R\{\bar x\}=R\big[ \sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\big].$$ We call the order of $\delta$ the maximum number of variables with same $x_i$ that appear multiplied together, for instance, $x_1x_2\sigma^3 x_2 +x_1x_2x_3$ has order $2$. By definition, a difference polynomial is zero if all the coefficients in $R$ of its monomials are zero.

Let $R\{\bar x,1\}$ be the set $$R\{\bar x,1\}=\left\{\displaystyle\sum_{0\leqslant i_1,\dots,i_n\leqslant n} r_{i_1,\dots,i_n}\sigma^{i_1}(x_1)\cdots\sigma^{i_n}(x_n):r_{\bar i}\in R,\ n\in\mathbb N\right\}$$ of those difference polynomials that are $n$-linear, i.e. of order $1$. In particular, one has $$R\{x,1\}=\left\{\displaystyle\sum_{i=0}^n r_{i}\sigma^{i}(x):\bar r\in R^n,\ n\in\mathbb N\right\},$$ which together with $+$ and $\circ$, defined as $$\left(\sum_{i=0}^n a_i {\sigma^i}\right)\circ\left(\sum_{j=0}^n b_j{\sigma^j}\right)=\sum_{i=0}^n\left(\sum_{j=0}^na_i\sigma^i(b_j) {\sigma^{i+j}}\right),$$ is a domain. We call the degree of $\delta\in R\{x,1\}$ the maximum $n$ such that $\sigma^n$ appears in $\delta$. For a difference polynomial $\delta$, we write $\delta_R$ the induced map on $R$.

Fact. If $R$ is a (possibly skew) field, then $(R\{x,1\},+,\circ)$ is left Euclidean.

Recall that $R$ is left-Ore if $Ra\cap Rb\neq (0)$ for all $a,b\in R\setminus\{0\}$, in which case $R^{-1}R$ is a division ring.

Corollary 1. If $R$ is a left-Ore domain, then $(R\{x,1\},+,\circ)$ is a left-Ore domain too.

Proof. If $R$ is left-Ore, then $R^{-1}R$ is a skew field, hence $R^{-1}R\{x,1\}$ is left-Euclidean, hence left-Ore, whence for every $a,b\in R\{x,1\}$, one has $R^{-1}R\{x,1\}a\cap R^{-1}R\{x,1\}b\neq (0)$, which implies $R\{x,1\}a\cap R\{x,1\}b\neq (0)$, so $R\{x,1\}$ is left-Ore.

Recall (?) that in a left-Ore domain $R$ (in particular in a commutative domain), there is a well-behaved notion of dimension (defined as the maximum cardinal of an $R$-independent family) for which the rank-nullity Theorem holds. In particular, if $f,g:M\rightarrow M$ are morphisms of an $R$-module $M$, then $dim Ker f\circ g\leqslant dim Ker f+dim Ker g$.

Lemma. Let $(R,+,\times,\sigma)$ be a difference left-Ore domain with fixed subdomain $F=\{x\in R:\sigma x=x\}$. Let $\delta\in R\{x,1\}\setminus\{0\}$ of degree $n$. The zero set of $\delta_R$ in $R$ is a $F$-module of dimension at most $n$.

Proof. By induction on $n$.

If $n=0$, then $\delta=r\in R\setminus\{0\}$ and has zero roots.

If $n=1$, then $\delta=\sigma-r id$. If $\delta$ has at least one root $a$, then $r=a^\sigma a^{-1}$. Every other root $b$ satisfies $a^\sigma a^{-1}=b^\sigma b^{-1}$ hence $\sigma(a^{-1}b)=a^{-1}b$ hence $a^{-1}b\in F$, whence $b\in aF$, so the zero set of $\delta_R$ has dimension $1$.

Induction step Assume $deg \delta=n+1$ and $\delta$ has at least one nonzero root $a\in R$. Applying Euclid division, in $(R^{-1}R\{x,1\},+,\circ)$, one has $\delta=\gamma\circ(\sigma-a^\sigma a^{-1}id)$ for some $\gamma\in R^{-1}R\{x,1\}$ of degree $n$. Then $r\gamma=\nu$ for some nonzero $r\in R$ and $\nu\in R\{x\}$ of degree $n$ and one conclude by induction, since $dim Ker\delta$ cannot exceed $dim Ker \nu+dim Ker(\sigma-a^\sigma a^{-1}id)$ and hence $n+1$ by induction hypothesis.

Corollary. Let K be a $\sigma$-field with fixed subfield $F$ and $[K:F]$ infinite, and $\delta\in K\{x_1,\dots,x_n,1\}\setminus \{0\}$ a $n$-linear difference polynomial. Then $\delta$ does not vanish on $K^n$.

Proof. By induction on $n$. For $n=1$, this is the previous Lemma. For the induction step, one can view $\delta$ as a linear twist in $x_n$ over $K\{x_1,\dots,x_{n-1},1\}$, where $K\{x_1,\dots,x_{n-1},1\}$ is a $\sigma$-domain with fixed subfield $F$ and also a left-Ore domain inductively by Corollary 1. By the above, the $x_n$-roots of $\delta$ form a $F$-module of finite dimension. In particular, since $[K:F]$ is infinite, there is $k\in K$ such tht $\delta(x_1,\dots,x_{n-1},k)$ is not the zero difference polynomial, so $\delta(\bar a,k)\neq 0$ for some $\bar a\in K^{n-1}$ by induction hypothesis.

Corollary. Let K be a $\sigma$-field with fixed subfield $F$ and $[K:F]$ infinite, and $\delta\in K\{x\}\setminus 0$ any nonzero difference polynomial. Then $\delta$ does not vanish on $K$.

Proof. If $\delta$ vanishes on $K$, then it is not linear by the above Corollary, hence $\delta(x+y)-\delta(x)-\delta(y)$ is nonzero, vanishes on $K^2$, and has order lower thab $\delta$. Iterating this method, we end on a nonero $n$-linear diffferential polynomial that vanishes on $K^n$, a contradiction with the above Corollary.

The answer is yes and appears in R. Cohn's book 'Difference algebra' Lemma II p. 201.

Let $(R,\sigma)$ be a difference domain (i.e. a domain equiped with a ring morphism $\sigma:R\rightarrow R$ satisfying $\sigma(a+b)=\sigma a+\sigma b$ and $\sigma(ab)=(\sigma a)(\sigma b$). Let $\bar x=(x_1,\dots,x_n)$. Let $R\{\bar x\}$ be the set of difference polynomials in variables $\bar x$, that is, the set of polynomials in variables $\{\sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\}$, $$R\{\bar x\}=R\big[ \sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\big].$$ We call the order of $\delta$ the maximum number of variables with same $x_i$ that appear multiplied together, for instance, $x_1x_2\sigma^3 x_2 +x_1x_2x_3$ has order $2$. By definition, a difference polynomial is zero if all the coefficients in $R$ of its monomials are zero.

Try for another proof by induction on the complexity of the differential equation:

Let $(R,\sigma)$ be a difference domain (i.e. a domain equiped with a ring morphism $\sigma:R\rightarrow R$ satisfying $\sigma(a+b)=\sigma a+\sigma b$ and $\sigma(ab)=(\sigma a)(\sigma b$). Let $\bar x=(x_1,\dots,x_n)$. Let $R\{\bar x\}$ be the set of difference polynomials in variables $\bar x$, that is, the set of polynomials in variables $\{\sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\}$, $$R\{\bar x\}=R\big[ \sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\big].$$ We call the order of $\delta$ the maximum number of variables with same $x_i$ that appear multiplied together, for instance, $x_1x_2\sigma^3 x_2 +x_1x_2x_3$ has order $2$. By definition, a difference polynomial is zero if all the coefficients in $R$ of its monomials are zero.

Let $(R,\sigma)$ be a difference domain (i.e. a domain equiped with a ring morphism $\sigma:R\rightarrow R$ satisfying $\sigma(a+b)=\sigma a+\sigma b$ and $\sigma(ab)=(\sigma a)(\sigma b$). Let $\bar x=(x_1,\dots,x_n)$. Let $R\{\bar x\}$ be the set of difference polynomials in variables $\bar x$, that is, the set of polynomials in variables $\{\sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\}$, $$R\{\bar x\}=R\big[ \sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\big].$$ We call the order of $\delta$ the maximum number of variables with same $x_i$ that appear multiplied together, for instance, $x_1x_2\sigma^3 x_2 +x_1x_2x_3$ has order $2$. By definition, a difference polynomial is zero if all the coefficients in $R$ of its monomials are zero.

Corollary. Let K be a $\sigma$-field with fixed subfield $F$ and $[K:F]$ infinite, and $\delta\in K\{x_1,\dots,x_n,1\}\setminus \{0\}$ a $n$-linear difference polynomial. Then $\delta$ does not vanish on $K^n$.

Proof. By induction on $n$. For $n=1$, this is the previous Lemma. For the induction step, one can view $\delta$ as a linear twist in $x_n$ over $K\{x_1,\dots,x_{n-1},1\}$, where $K\{x_1,\dots,x_{n-1},1\}$ is a $\sigma$-domain with fixed subfield $F$ and also a left-Ore domain inductively by Corollary 1. By the above, the $x_n$-roots of $\delta$ form a $F$-module of finite dimension. In particular, since $[K:F]$ is infinite, there is $k\in K$ such tht $\delta(x_1,\dots,x_{n-1},k)$ is not the zero difference polynomial, so $\delta(\bar a,k)\neq 0$ for some $\bar a\in K^{n-1}$ by induction hypothesis.

The answer is yes and appears in R. Cohn's book 'Difference algebra' Lemma II p. 201.

Let $(R,\sigma)$ be a difference domain (i.e. a domain equiped with a ring morphism $\sigma:R\rightarrow R$ satisfying $\sigma(a+b)=\sigma a+\sigma b$ and $\sigma(ab)=(\sigma a)(\sigma b$). Let $\bar x=(x_1,\dots,x_n)$. Let $R\{\bar x\}$ be the set of difference polynomials in variables $\bar x$, that is, the set of polynomials in variables $\{\sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\}$, $$R\{\bar x\}=R\big[ \sigma^j(x_i):j\in\mathbf N,\ i=1,\dots,n\big].$$ We call the order of $\delta$ the maximum number of variables with same $x_i$ that appear multiplied together, for instance, $x_1x_2\sigma^3 x_2 +x_1x_2x_3$ has order $2$. By definition, a difference polynomial is zero if all the coefficients in $R$ of its monomials are zero.

Corollary. Let K be a $\sigma$-field with fixed subfield $F$ and $[K:F]$ infinite, and $\delta\in K\{x_1,\dots,x_n,1\}\setminus \{0\}$ a $n$-linear difference polynomial. Then $\delta$ does not vanish on $K^n$.

Proof. By induction on $n$. For $n=1$, this is the previous Lemma. For the induction step, one can view $\delta$ as a linear twist in $x_n$ over $K\{x_1,\dots,x_{n-1},1\}$, where $K\{x_1,\dots,x_{n-1},1\}$ is a $\sigma$-domain with fixed subfield $F$ and also a left-Ore domain inductively by Corollary 1. By the above, the $x_n$-roots of $\delta$ form a $F$-module of finite dimension. In particular, since $[K:F]$ is infinite, there is $k\in K$ such tht $\delta(x_1,\dots,x_{n-1},k)$ is not the zero difference polynomial, so $\delta(\bar a,k)\neq 0$ for some $\bar a\in K^{n-1}$ by induction hypothesis.