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Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;g(n)=h(\phi(n)),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}\;$ with $k>0$.

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|\ge b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$ Note that $|f(t)|\ge b\;$ is implied by $f(t)$ being a nonzero integer.

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.

Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;h(\phi(n))=g(n),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}\;$ with $k>0$.

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|\ge b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$ Note that $|f(t)|\ge b\;$ is implied by $f(t)$ being a nonzero integer.

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.

Added note: I implicitly define $h$ by $h(n)=g((n-1)/x)$ only for $n=1\pmod{x}$ and this is equivalent to $h(\phi(n))=g(n).$

Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;g(n)=h(\phi(n)),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}.$

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|\ge b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$ Note that $|f(t)|\ge b\;$ is implied by $f(t)$ being a nonzero integer.

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.

Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;g(n)=h(\phi(n)),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}\;$ with $k>0$.

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|\ge b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$ Note that $|f(t)|\ge b\;$ is implied by $f(t)$ being a nonzero integer.

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.

Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;g(n)=h(\phi(n)),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}.$

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|>b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.

Let $\;g(n):=nf(n).\;$ If $x=0$ then the functional equation is $\;g(n)+g(m)=g(n+m)\;$ for all $n,m\in\mathbb{Z},\;$ which is Cauchy's functional equation for integers and the general solution is $g(n)=cn$ which implies $\;c=f(n)\;$ for all $\;n\ne 0, n\in\mathbb{Z},\;$ while $f(0)$ is arbitrary.

Now assume $x\ne 0,\;\phi(n):=1+xn,\;g(n)=h(\phi(n)),\;$ and let $n\oplus m:=n+m+xnm\;$ where $\phi(n\oplus m)=\phi(n)\phi(m).\;$ Then the functional equation is $\;g(n)+g(m)=g(n\oplus m)\;$ for all $n,m\in\mathbb{Z},\;$ but now rewriting it as $\; h(\phi(n))+h(\phi(m))=h(\phi(n)\phi(m))\;$ leads to $\;k\:h(\phi(n))=h(\phi(n)^k)\;$ for all $k,n,m\in\mathbb{Z}.$

Given $n\ne0,\;$let $\;t:=(\phi(n)^k-1)/x.\;$ Now $\;g((n-1)/x)=h(n),\;$ thus $\;k\:n\:f(n)=t\:f(t).$ Now, assuming that $\;|f(t)|\ge b>0\;$ for $k$ big enough, then the right side grows exponentially and the left linearly which is a contradiction. Thus eventually $f(t)=0$ and hence $f(n)=0.$ Note that $|f(t)|\ge b\;$ is implied by $f(t)$ being a nonzero integer.

My proof is very similar to the one in the other answer, but has a few more details. I think Forsyth's remark may be similar to the one in that too narrow margin of Fermat.