http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78454 says even a bit more (it's not hard to get out of this that only the linear polynomials with rational coefficients work).

$ax+b$, with $a,b\in \mathbb{Q}$, $a\neq 0$.

This is a beautiful problem. First, I claim that all such polynomials have rational coefficients. This isn't too bad: Say that the polynomial was $f$, and $\deg f = j$. Now apply Lagrange Interpolation on any $j+1$ rational points, and we're good for this part. Now say that $\deg f \ge 2$. I claim that $f$ passes through a mixed point. First of all, multiply all the coefficients of $f$ by a suitable integer so $f$ becomes an integer polynomial. Subtract $f(0)$ from $f$, so $f(0) = 0$. Now take $-f$ if $f$ had a negative leading coefficient. For what follows, consider the domain $x > A$, where $A$ is the largest root of $f'(x)$. It is clear that $f$ is monotonic and hence bijective in this domain. Now let $a$ be the leading coefficient of $f$. Assume that $f$ passes through no mixed points. Then, it must be true that for all intergal $c$ larger than $f(A)$, there exists a unique $x > A$ such that $f(x) = c$. Moreover, by the hypothesis, this $x$ must be rational. In fact, by the rational root theorem, $x$ is a integer multiple of the fraction $\dfrac{1}{a}$! Let the sequence $k_c$ for all integers $c > A$ be this exact $x$. Since $f(x)$ is strictly increasing over this domain, so must $k_c$. Observe that \[ k_{c+1} - k_c \ge \frac{1}{a} \]since they are obviously different, and $k_{c+1} > k_c$. Also observe that \[ (c+1) - c = 1.\]By the mean value theorem, there exists a $u$ in the domain $[k_c, k_{c+1}]$ such that \[ f'(u) = \frac{f(k_{c+1}) - f(k_c)}{k_{c+1}-k_c} = \frac{1}{k_{c+1}-k_c} \le a\] However, since the degree of $f$ is larger than $1$, and $k_c$ tends to infinity, $f'$ is also monotonically increasing past a certain point. So we just need to take a sufficiently large $c$ to get contradiction in the above equation. So the only polynomials which have a chance of working are $f(x) = ax + b$, where $a, b \in \mathbb{Q}$. But it is trivial that these polynomials work, so we're done.

Quite a nice problem. Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\dots +a_{1}x+a_{0}$ be a polynomial which goes through no mixed points. The key claim is the following Lemma. $P$ must only have rational coefficients. Proof: We'll only use that $P(\mathbb{N})\in\mathbb{Q}$ as otherwise $P$ would go through a mixed point. Recall that if $g\in\mathbb{R}[x]$, $\Delta_t g(n):=g(n+t)-g(n)$ and $D_k(g)=\Delta_{2^{k-1}}(D_{k-1}(f))$ where $D_0(g)=g$, then $\deg(\Delta_{k+1})+1=\deg(\Delta_{k})$ for $0\leq k\leq \deg g$ and $D_{\deg g+1}(g)$ is the zero polynomial. Hence, since $P(n)\in\mathbb{Q}$ for all $n\in\mathbb{N}$ we have that: \[\mathbb{Q}\ni D_{n}(P)=D_{n}\left(\sum\limits_{i=0}^{n} a_{i}x^{i}\right)=\sum\limits_{i=0}^{n} D_{n}\left(a_{i}x^{i}\right)=D_{n}(a_{n}x^{n})=c\cdot a_{n}\]where $c$ is some nonzero constant as $D_{n}(a_{n}x^{n})$ is a constant polynomial, but not the constant zero. Also, we know that: \[c=\frac{1}{a_{n}}D_{n}(P)=\frac{1}{a_{n}}\sum\limits_{k=1}^{2^{n}} \varepsilon_{k} P(x+k)=\sum\limits_{k=1}^{2^{n}} \varepsilon_{k} (x+k)^{n}\Longrightarrow c = \sum\limits_{k=1}^{2^{n}} \varepsilon_{k} k^{n} \in \mathbb{Z}\]where $\varepsilon_{k}\in\{+1,-1\}$ and depends on $D_{n}(P)$. Therefore $c\cdot a_{n}\in \mathbb{Q}$ and $c\in \mathbb{Z}\Longrightarrow a_{n}\in\mathbb{Q}$. Now, we must have $P_{1}(\mathbb{Q})\in \mathbb{Q}$ where $P_{1}(x)=P(x)-a_{n}x^{n}$ and applying the same procedure implies that $a_{n-1}\in\mathbb{Q}$, $a_{n-2}\in\mathbb{Q}$ and so on, we get inductively that every coefficient must be rational. If $P$ is a constant polynomial, then all inputs of $x$ spit out a rational, hence there are mixed points $P$ goes through and if $P$ is linear, then it doesn't go through any mixed points as $P(x)=a_{1}x+a_{0}$ for $a_{1}, a_{0}\in\mathbb{Q}$ and $a_{1}\neq 0$ as $P(x)\in\mathbb{Q}\iff x\in\mathbb{Q}$. We'll show that if $\deg P\geq 2$, then $P$ can't give irrational outputs for all irrational inputs. WLOG we can assume that $P$ has integer coefficients because we can multiply the polynomial by some $C\in\mathbb{N}$ such that $Ca_{i}\in\mathbb{Z}$ $\forall i$ and this wouldn't change if $P$ goes through any mixed points. WLOG let's also assume that $a_{n}>0$ (if that's not the case just negate the polynomial). Now we must have that $P(x)-n=0$ only has rational roots for all integers $n$. Let $n=p+a_{0}$ for any prime $p>\max\{a_{n}, |P(0)|+|a_{0}|\}$. Since $P(0)-p-a_{0}<0$ and $P(x)-p-a_{0}>0$ for a large enough $x$ (as the leading coefficient of $P$ is positive), $P(x)-p-a_{0}=0$ has a real solution due to IVT and this root must be rational to avoid contradiction with the assumption that $P$ doesn't go through mixed points. Now, by the Rational Root Theorem, it follows that if this root is $\frac{b}{c}$ where $b,c\in\mathbb{Z}$ and $\gcd(b,c)=1$, then $b\mid p$ and $c\mid a_{n}$ since $-p$ and $a_{n}$ are respectively the constant term and leading coefficient of $P(x)-p-a_{0}=0$. Note that the two equations $P(x)-p_{1}-a_{0}=0$ and $P(x)-p_{2}-a_{0}=0$ clearly can't share a root as that would imply that $p_{1}=p_{2}$. Now $\frac{b}{c}$ is either of the form $\frac{1}{d}$ where $d\in\mathbb{Z}$ and $d\mid a_{n}$ or $\frac{p}{d}$ where $d\in\mathbb{Z}$ and $d\mid a_{n}$. Since there are finitely many rational numbers of the first form we must have that for infinitely many primes $p$, $\frac{b}{c}=\frac{p}{d}$. Now, as there are finitely many choices for $d$, it follows that for infinitely many primes $p$ we have that $\frac{b}{c}=\frac{p}{D}$ for a fixed integer $D$, which divides $a_{n}$. Hence $P\left(\frac{p}{D}\right)-p-a_{0}=0$ for infinitely many primes $p$. This implies that the polynomial $P(X)-DX-a_{0}=0$ has infinitely many roots and is therefore the zero polynomial, but $\deg(P)\geq 2$, contradiction.