Are there smooth affine varieties which cannot be embedded into smooth projective varieties? I just asked a similar question here: Do there exist manifolds which cannot be smoothly embedded in a compact manifold?
The answer there was pretty topological, so I think this merits its own question. (Though it would be interesting to me if a similar construction could be done in the algebraic category.)
Let $M$ be a smooth affine variety (over an algebraically closed field). Does there exist a smooth projective variety $N$ into which it embeds as an open subset?
 A: It is already mentioned by Mohan in the comments, but the problem is essentially equivalent to resolution of singularities.
If we have resolution (e.g. if $\operatorname{char} k = 0$, or $\dim X \leq 2$), then we can just start with any projective model (e.g. take an affine model and take the closure in the associated projective space). Blowing up will make the variety smooth. (Note: it's not obvious that you can do this without affecting the given variety $X$. You would need to dive into the proof to see whether you may assume that you don't change the stuff that is already smooth. I have not done this.)
On the other hand, if you could prove your statement, then a weak version of resolution would follow. Indeed given any $X$ over $k$ projective, let $X^0$ be the smooth locus (which is nonempty when $k$ is perfect, e.g. algebraically closed). Then your result implies that we can find $Y \supseteq X^0$ smooth projective. This at least gives
$$\phi \colon Y \dashrightarrow X$$
birational from the smooth variety $Y$ to $X$. The usual way to extend to an actual morphism from a smooth projective onto $X^0$ is by resolving the graph of $\phi$, which we cannot do (unless we already know resolution).
However, it seems to me that proving your statement could well be a major breaktrough already! (I hope someone will correct me if I'm wrong.)
