You mentioned that for finite fields, such f exist. Under the following assumption, I believe there is an infinite field with the same such that an f exists. (My field theory is very weak, so I'm not sure how obviously correct or obviously incorrect these assumptions are):

There is a sequence of finite fields $F_q$ with appropriate polynomials $f_q$ such that

i) $q< p$ implies $F_q$ has fewer elements thatn $F_p$

ii) deg$(f_{q}) \leq n\in\mathbb{N}$ (uniformly bounded)

iii) the number of points missed by $f_q$ is uniformly bounded above by $m\in\mathbb{N}$

Under these assumptions, construct an infinite field as follows:

Let F be the set of usual field axioms (expressed in first order logic).

Let $\psi$ be the first order statement "there are coefficients $a_{0}$ through $a_{n}$ and there are other points $y_{1}$ through $y_{m}$ such that for any $x$ we have $a_{n} x^{n} + ... + a_{1}x + a_{0} \neq y_{k}$ for any $k$ and for all $w$ which are not equal to $y_{1}$ through $y_{m}$ there is a $x$ such that $a_{n}x^{n} + ... + a_{0} = w$"

More colloquially, $\psi$ says "the polynomial $f(x) = a_{n}x^{n} + ... + a_{0}$ misses $y_1$ through $y_m$ but nothing else"

(One can, e.g., set $a_{n} = 0$ or $y_{1} = y_{2}$ if for a given finite field, the degree is smaller or $f_q$ misses fewer points)

Let $\phi_k$ be the first order statement "There are at least $k$ elements" (i.e., there exist $x_{1}$ through $x_{k}$ such that they are pairwise nonequal).

Finally, set $T = F \cup {\psi} \cup {\phi_{n}}$.

A model of $T$ is simply a set with interpretations for everything such that all the statements of $T$ are satisfied. In other words, a model is a field (because is satisfies F) which is infinite (because it simultaneously satisfies all of the $\phi_n$) which has a polynomial like you want (because of $\psi$).

Godel's completeness theorem says that $T$ has a model iff $T$ is consistent. The compactness theorem for first order logic says that $T$ is consistent iff every finite subset of $T$ is consistent.

Hence, by applying Godel's completeness theorem again, we need only show that every finite subset of $T$ has a model.

Choosing a finite $T_{0}\subseteq T$, we may, without loss of generality, enlarge it by including $F$ and $\psi$ because a model of $T_{0}\cup F\cup \psi$ will model $T_{0}$.

Now, since $T_{0}$ is finite, there is a largest $N$ such that $\phi_{N}$ is in $T_{0}$. Because of this, a model of $T_{0}$ is simply a finite field of at cardinality at least $N$ with a choice of function $f$ satisfying what you want (with bound on deg(f) and the number of points missed in the image). But the existence of such a field was precisely the assumption made at the top of the post.

Now, hopefully someone can shed some light as to whether or not the assumption is true.

You mentioned that for finite fields, such f exist. Under the following assumption, I believe there is an infinite field with the same such that an f exists. (My field theory is very weak, so I'm not sure how obviously correct or obviously incorrect these assumptions are):

There is a sequence of finite fields $F_q$ with appropriate polynomials $f_q$ such that

i) $q_{1}< q_{2}$ implies $F_{q_{1}}$ has fewer elements than $F_{q_{2}}$ (Edited notation a bit here)

ii) deg$(f_{q}) \leq n\in\mathbb{N}$ (uniformly bounded)

iii) the number of points missed by $f_q$ is uniformly bounded above by $m\in\mathbb{N}$

Under these assumptions, construct an infinite field as follows:

Let F be the set of usual field axioms (expressed in first order logic).

Let $\psi$ be the first order statement "there are coefficients $a_{0}$ through $a_{n}$ and there are other points $y_{1}$ through $y_{m}$ such that for any $x$ we have $a_{n} x^{n} + ... + a_{1}x + a_{0} \neq y_{k}$ for any $k$ and for all $w$ which are not equal to $y_{1}$ through $y_{m}$ there is a $x$ such that $a_{n}x^{n} + ... + a_{0} = w$"

More colloquially, $\psi$ says "the polynomial $f(x) = a_{n}x^{n} + ... + a_{0}$ misses $y_1$ through $y_m$ but nothing else"

(One can, e.g., set $a_{n} = 0$ or $y_{1} = y_{2}$ if for a given finite field, the degree is smaller or $f_q$ misses fewer points)

Let $\phi_k$ be the first order statement "There are at least $k$ elements" (i.e., there exist $x_{1}$ through $x_{k}$ such that they are pairwise nonequal).

Finally, set $T = F \cup {\psi} \cup {\phi_{n}}$.

A model of $T$ is simply a set with interpretations for everything such that all the statements of $T$ are satisfied. In other words, a model is a field (because is satisfies F) which is infinite (because it simultaneously satisfies all of the $\phi_n$) which has a polynomial like you want (because of $\psi$).

Godel's completeness theorem says that $T$ has a model iff $T$ is consistent. The compactness theorem for first order logic says that $T$ is consistent iff every finite subset of $T$ is consistent.

Hence, by applying Godel's completeness theorem again, we need only show that every finite subset of $T$ has a model.

Choosing a finite $T_{0}\subseteq T$, we may, without loss of generality, enlarge it by including $F$ and $\psi$ because a model of $T_{0}\cup F\cup \psi$ will model $T_{0}$.

Now, since $T_{0}$ is finite, there is a largest $N$ such that $\phi_{N}$ is in $T_{0}$. Because of this, a model of $T_{0}$ is simply a finite field of at cardinality at least $N$ with a choice of function $f$ satisfying what you want (with bound on deg(f) and the number of points missed in the image). But the existence of such a field was precisely the assumption made at the top of the post.

Now, hopefully someone can shed some light as to whether or not the assumption is true.