Let's assume that we have a flat object which has a random shape and does not have a symmetry axis. Let's assume the flat object lays in the xy-plane.
Any rigid body in principle has a set of three principal axes, which can be found generally by diagonalizing the matrix corresponding to the inertia tensor.
For this specific object, do the 2 (of the 3) principal axes have to be necessarily in the xy-plane as well? I could not find any general theorem regarding this.
Consider a small particle of mass $m_i$ located at $(x_i,y_i)$ on the flat body.
The contribution of this particle to the body MMOI tensor about the origin is
$$ {\rm I} = \sum_i m_{i}\begin{bmatrix}y_{i}^{2} & \text{-}x_{i}y_{i}\\ \text{-}x_{i}y_{i} & x_{i}^{2}\\ & & x_{i}^{2}+y_{i}^{2} \end{bmatrix} $$
with the missing elements above being zero.
In total the general structure of the MMOI tensor is
$$ {\rm I}=\begin{bmatrix}I_{xx} & I_{xy}\\ I_{xy} & I_{yy}\\ & & I_{zz} \end{bmatrix} $$
Now you are asking about the eigenvectors for this matrix, and if it possible to have one that contains components on both z and xy directions.
There are three eigenvalues for this matrix
$$\begin{aligned}\lambda_{1} & =\frac{I_{xx}+I_{yy}+\sqrt{\left(I_{xx}-I_{yy}\right)^{2}-4I_{xy}}}{2} \\ \lambda_{2} & =\frac{I_{xx}+I_{yy}-\sqrt{\left(I_{xx}-I_{yy}\right)^{2}-4I_{xy}}}{2} \\ \lambda_{3} & =I_{zz}\end{aligned}$$
The first two correspond to eigenvectors that are in-plane (only have xy components) and the third one to an eigenvalue with no xy components and only a z component.
You can confirm this last part readily with $$\left(\begin{bmatrix}I_{xx} & I_{xy}\\ I_{xy} & I_{yy}\\ & & I_{zz} \end{bmatrix}-\lambda_{3}\right)\begin{pmatrix}0\\ 0\\ 1 \end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix}$$
and you can show that exists such angles $\theta_1$ and $\theta_2$ to also solve the eigenvector problem for the other two eigenvalues
$$\small \begin{aligned}\left(\begin{bmatrix}I_{xx} & I_{xy}\\ I_{xy} & I_{yy}\\ & & I_{zz} \end{bmatrix}-\lambda_{1}\right)\begin{pmatrix}\cos\theta_{1}\\ \sin\theta_{1}\\ 0 \end{pmatrix} & =\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix} & \left(\begin{bmatrix}I_{xx} & I_{xy}\\ I_{xy} & I_{yy}\\ & & I_{zz} \end{bmatrix}-\lambda_{2}\right)\begin{pmatrix}\cos\theta_{2}\\ \sin\theta_{2}\\ 0 \end{pmatrix} & =\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix}\end{aligned}$$
But is there a special case with an eigenvector of the form
$$\left(\begin{bmatrix}I_{xx} & I_{xy}\\ I_{xy} & I_{yy}\\ & & I_{zz} \end{bmatrix}-\lambda_{i}\right)\begin{pmatrix}\cos\theta\\ \sin\theta\\ e_{z} \end{pmatrix}=\begin{pmatrix}0\\ 0\\ 0 \end{pmatrix}$$
which means the principal axis comes of the plane but it is not parallel to the z axis. There is only one special circumstance that this can happen with
$$ I_{xy}=\pm\sqrt{I_{zz}-I_{xx}}\sqrt{I_{zz}-I_{yy}} $$
I am not sure what shape can produce this exact value, and if it physically possible, but in theory it might be possible to produce such as shape. In practice no, because you can never make something with exact dimensions.
The matrix of inertia is symmetric, and all symmetric matrices have real eigenvalues and orthogonal eigenvectors, as shown here.
As one of the axis (eigenvectors) of inertia must of course be parallel to the z-axis, the other 2 must be in the xy axis in order to result in a set of 3 orthogonal axis (eigenvectors).
