I'd like to suggest that this isn't quite the right question. At least, it seems to me that modifying the question (in a direction that Theo was hinting) would be more interesting.

The problem with the question as asked is that, for a given category C, the mere existence of a faithful functor C --> Set tells you very little indeed. Perhaps you have some reason for wanting to know that I can't see. But a condition that seems to have more bite is 'small-concreteness', defined as follows.

Let C be a category. A set-valued functor U: C --> Set is small if it can be expressed as a small colimit of representables. Call a category C small-concrete if there exists a small, faithful functor C --> Set. In the special case that C is small, all set-valued functors on C are small and small-concrete = concrete.

It's not too hard to show that a category is small-concrete if and only if it admits a generating set. (A generating set in a category C is a [small] set S of objects such that, for any distinct maps f, g: a --> b in C, there exist s in S and q: s --> a such that fq \neq gq.) The existence of a generating set is one of the conditions in the Special Adjoint Functor Theorem: see Categories for the Working Mathematician.

You can exploit this as follows. Suppose you want to show that the category of affine schemes is not small-concrete (which would imply that the category of all schemes isn't either). Assuming for a contradiction that it is small-concrete, the category Ring of commutative rings has a cogenerating set. Since Ring is locally small and small-complete, the Special Adjoint Functor Theorem tells us that every limit-preserving functor from Ring to a locally small category has a left adjoint. I guess it's possible to cook up (or look up) an example of a limit-preserving functor out of Ring that doesn't have a left adjoint. That would produce the desired contradiction.

I'd like to suggest that this isn't quite the right question. At least, it seems to me that modifying the question (in a direction that Theo was hinting) would be more interesting.

The problem with the question as asked is that, for a given category $C$, the mere existence of a faithful functor $C \to \mathbf{Set}$ tells you very little indeed. Perhaps you have some reason for wanting to know that I can't see. But a condition that seems to have more bite is 'small-concreteness', defined as follows.

Let C be a category. A set-valued functor $U: C \to \mathbf{Set}$ is small if it can be expressed as a small colimit of representables. Call a category $C$ small-concrete if there exists a small, faithful functor $C \to \mathbf{Set}$. In the special case that $C$ is small, all set-valued functors on $C$ are small and small-concrete = concrete.

It's not too hard to show that a category is small-concrete if and only if it admits a generating set. (A generating set in a category $C$ is a [small] set $S$ of objects such that, for any distinct maps $f, g: a \to b$ in $C$, there exist $s \in S$ and $q: s \to a$ such that $fq \neq gq$.) The existence of a generating set is one of the conditions in the Special Adjoint Functor Theorem: see Categories for the Working Mathematician.

You can exploit this as follows. Suppose you want to show that the category of affine schemes is not small-concrete (which would imply that the category of all schemes isn't either). Assuming for a contradiction that it is small-concrete, the category $\mathbf{Ring}$ of commutative rings has a cogenerating set. Since $\mathbf{Ring}$ is locally small and small-complete, the Special Adjoint Functor Theorem tells us that every limit-preserving functor from $\mathbf{Ring}$ to a locally small category has a left adjoint. I guess it's possible to cook up (or look up) an example of a limit-preserving functor out of $\mathbf{Ring}$ that doesn't have a left adjoint. That would produce the desired contradiction.