When does QCoh have 'enough perfect complexes'? Let $X$ be a derived fpqc stack on the $\infty$-category of connective spectral affine schemes $\mathbf{Aff}^{\mathrm{cn}}=(\mathbf{Ring}^{\mathrm{cn}}_{E_\infty})^{\mathrm{op}}$, that is to say, a functor $X:(\mathbf{Aff}^{\mathrm{cn}})^{\mathrm{op}}\to \mathcal{S}$ satisfying fpqc descent.  Then we can define its $\infty$-category of quasicoherent sheaves formally by a Kan extension. 
Say that a symmetric monoidal stable $\infty$-category $\mathcal{C}$ 'has enough perfect objects' if its full subcategory of dualizable objects is dense (that is to say the induced functor $\mathcal{C}\to \operatorname{Ind}(\operatorname{Perf}(\mathcal{C}))$ is fully faithful). 
Are there examples of fpqc stacks $X$ as above for which $\operatorname{QCoh}(X)$ does not have enough perfect objects?
What about if we restrict our question to ((Quasi)-Geometric Stacks, Artin Stacks, Deligne-Mumford Stacks, Algebraic Spaces, Schemes)?  
For certain, this is true for quasicompact quasiseparated schemes and algebraic spaces, as well as quasi-Geometric stacks $X$ such that $\operatorname{QCoh}(X)$ is compactly generated and the structure sheaf is a compact object (proven in Lurie, Spectral Algebraic Geometry).  
 A: Robert Thomason was the first person to draw attention to this question, before derived schemes and infinity categories. I believe that he proved that for a quasi-compact and quasi-separated scheme that $D_{qc}=\textrm{Ind}(\textrm{Perf})$. For example, see Thomason-Trobaugh section 2.3, though at first glance it appears that only proves the weaker statement that it has enough perfect complexes.
Somewhere he gives two examples to show the necessity of the two hypotheses. Consider an affine scheme with a point of infinite codimension, say, $X=\textrm{Spec}\;k[x_1,x_2,…]$. Let $U$ be the complement of the point. It is not quasi-compact. Let $Y=X\cup_U X$ be $X$ with the point doubled. It is not quasi-separated. A perfect complex is built from finitely many operations, so its support has finite codimension, so they do not notice the points of infinite codimension, so the three schemes all have the same perfect complexes. But they have different quasi-coherent complexes, such as the skyscrapers on the origins. In particular, $Y$ has two such sheaves, but they cannot be distinguished by perfect complexes. Whereas $U$ has too few sheaves, so it fails the strong hypothesis of equivalence of categories, but it still has enough perfects: $\textrm{QCoh}(U)\subset \textrm{QCoh}(X)=\textrm{Ind}(\textrm{Perf(X)})=\textrm{Ind}(\textrm{Perf(U)})$

Is there a finite dimensional example? For example, consider this non-quasi-compact scheme built from varieties. Let $Z_0=\mathbb A^2$ and $x_0=0\in Z_0$. Let $Z_{n+1}$ be the blow up of $Z_n$ at $x_n$ and let $x_{n+1}$ be a point in the exceptional fiber. Let $U_n=Z_n-\{x_n\}$. Then $U_n$ is open in $U_{n+1}$ and let $U'=\bigcup U_n$. Does it satisfy $D_{qc}=\textrm{Ind}(\textrm{Perf})$? I believe that it can be compactified by adding a 2-dimensional valuation ring. If so, we could double that point to get a non-quasi-separated scheme. Would it fail to have enough perfects?
