What is needed to force polynomial-ring automorphisms to be affine? Is there an integral domain $R$ and a polynomial-ring automorphism $\: \phi : R[x] \to R[x] \:$ such that, for $\: i : R\to R[x] \:$ the canonical embedding, $\;\;\; \phi \circ i \: = \: i \;\;$ and

there do not exist elements $c$ and $u$ of $R$ such that $u$ is a unit and $\;\; \phi(x) \: = \: c+(u\hspace{-0.04 in}\cdot \hspace{-0.04 in}x) \;\;\;$?

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More generally, what properties of rings $R$ suffice to rule out the existence of such a $\phi$?
Clearly, $R$ must have a $1$, since otherwise the identity function would be such a $\phi$.

If $R$ has a $1$ and an $\epsilon$ such that $\: \epsilon^{\hspace{.02 in}2} \hspace{-0.0 in} = 0 \neq \epsilon \:$ and $\: \epsilon \in \:$Z$(R)\:$, $\:$ then the homomorphism given by,

for $\: c\in R \:$, $\:$ $\: i(c) \mapsto i(c) \:$ and $\;\; x \: \mapsto \: x+(\epsilon \hspace{-0.04 in}\cdot \hspace{-0.04 in}x^{\hspace{.02 in}2}) \;\;$, $\;\;\;$ is such a $\phi$.

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 A: In light of our conversation, let me change my answer as follows:
Let us assume that $R$ is commutative. If $R$ is not commutative other nastiness will follow.
If $R$ is commutative, the property that $R$ is fixed by the automorphism is equivalent to $\phi: R[x] \to R[x]$ being an $R$-algebra automorphism. This means that generators must map to generators under $\phi$. Fixing a generator $\langle x \rangle = R[x]$ for the domain, means that it is sufficient to find the collection of all generators for $R[x]$, define a set-theoretic map $x \mapsto g$ for some generator $g$, then extend "linearly/multiplicatively." 
Thus the question is reduced to, "Under what conditions on $R$ allows for more than just  'affine' generators $p(x) = a+bx$ of $R[x]$?" Now, I think it is reasonable to say that if $R$ is an integral domain (so it has no zero divisors), that this is not possible. The reason being that if $R$ is an integral domain, the polynomial degree map does what we think it should, i.e. $\deg p(x)q(x) = \deg p(x) + \deg q(x)$, so if $g(x)$ is a generator of degree at least $2$, we will not be able to generate $R[x]$. This technically resolves your question for integral domains.
The more interesting question is, what if we do not limit ourselves to integral domains? Your example demonstrates that such an automorphism is possible if the ring admits nilpotent elements. In fact, your example should generalize to more than just square-nilpotent elements. Namely, if there exists $\epsilon \in R\setminus\{0\}$ such that $\epsilon^p = 0$ then the map on generators sending $x \mapsto x + \epsilon^{p-1} x^p$ will be such an automorphism, since we can generate $x$ via $\phi(x-\epsilon^{p-1} x^p)$ and all other powers of $x^n$ are derivative of that. We can generalize further, by saying that if $R$ admits a family of nilpotents $\epsilon_i$ such that $\epsilon_i^{p_i}=0$ then $x \mapsto x + \sum_{i\in I}\epsilon_i^{p_i-1} x^{p_i}$ should be a generator for any finite index set $I$.
Thus I suppose the answer to your question is: If $R$ is commutative, such automorphism exist so long as the nilradical $\mathfrak n \subset R$ is non-trivial.
In the non-commutative case, you can no longer say that $\phi$ is an $R$-algebra automorphism, so it is not clear to me that we can make such an assertion on generators.
