Are there Zariski-continuous maps between algebraic sets that are not polynomial maps?
If I understand your question correctly, then (1) yes, there are going to be lots of Zariski continuous maps that are not polynomial (although, if one asks that all higher products of the map preserve the Zariski topology on the corresponding sources and targets, this is a much stronger restriction, I think --- does anyone know the details?); and (2) the usual extra structure that one adds is the sheaf of regular functions on source and target; polynomial maps pullback regular functions on the target to regular functions on the source. Thus varieties are naturally thought of as being objects in the category of locally ringed spaces (i.e. spaces equipped with a certain structure sheaf of rings) rather than just in the category of topological spaces.
Just an easy remark. If $k = \mathbb{F}_q$ is the finite field with $q$ elements, then every function $k^n \to k$ is a polynomial map. In fact, for every point $p = (a_1, \dots, a_n) \in k^n$ consider the polynomial $$ f(x_1, \dots, x_n) = (-1)^n \prod_{i=1}^n \prod_{b \in \mathbb{F}_q \setminus \{ a_i \} } (x_i - b); $$ $f(p) = 1$ and $f$ is zero in $k^n \setminus \{ p \}$.
The complex conjugation $\bar \cdot : \Bbb C \to \Bbb C$ is a concrete example of a Zariski-continuous function that is not a polynomial function.
It is not a polynomial function because if it were, since this property is independent of the underlying topology, $\bar \cdot$ would also be a polynomial function in the transcendental topology (the "usual" topology on $\Bbb C$), therefore it would also be holomorphic - which it clearly is not, since it doesn't satisfy the Cauchy-Riemann relations.
To see that it is continuous, notice that if $Z = f^{-1}(\{0\})$ with $f \in \Bbb C[x]$, and if $\bar f$ is the polynomial obtained by conjugating the coefficients of $f$, then
$$\bar \cdot ^{-1} (Z) = \bar \cdot (Z) = \{ \bar x \in \Bbb C \mid f(x) = 0\} = \{ \bar x \in \Bbb C \mid \overline{f(x)} = 0\} = \{ \bar x \in \Bbb C \mid \bar f(\bar x) = 0\} = \\ \{ z \in \Bbb C \mid \bar f(z) = 0\} = \bar f ^{-1} (\{0\})$$
which is Zariski-closed.
If $C$ is an arbitrary Zariski-closed set, then there exist polynomials $f_1, \dots, f_k$ such that
$$C = f_1 ^{-1} (\{0\}) \cap \dots \cap f_k ^{-1} (\{0\}) ,$$
whence it follows that
$$\bar \cdot ^{-1} (C) = \bar \cdot ^{-1} \Big( f_1 ^{-1} (\{0\}) \cap \dots \cap f_k ^{-1} (\{0\}) \Big) = \bar f_1 ^{-1} (\{0\}) \cap \dots \cap \bar f_k ^{-1} (\{0\})$$ which, being an intersection of Zariski-closed sets, is itself Zariski-closed.
Since the preimage of every Zariski-closed set is Zariski-closed, it follows that $\bar \cdot$ is Zariski-continuous.