No, any POVM, including POVMs whose elements do not have unit rank, can be via interpreted Naimark's theorem as a projective measurement in a higher-dimensional space.
Derivation of the dilated representation
More precisely, let $\{\mu_a\}_a\subset\operatorname{Pos}(\mathbb{C}^d)$ be a POVM, so that $\sum_a\mu_a=I$.
Define the linear operator $V:\mathbb{C}^n\to\mathbb{C}^n\otimes\mathbb{C}^m$ as
$$Vu = \sum_a (\sqrt{\mu_a}u)\otimes |a\rangle, \qquad u\in\mathbb{C}^n,$$
for some orthonormal basis $\{|a\rangle\}_a$ for $\mathbb{C}^m$. You can verify that this $V$ is an isometry, and its Hermitian conjugate acts on basis vectors as
$$V^\dagger(v\otimes w) = \sum_a \langle w,a\rangle \sqrt{\mu_a} v, \qquad v\in\mathbb{C}^n,\, w\in\mathbb{C}^m.$$
This expression can be obtained from the definition of Hermitian conjugate via the inner product:
$$\langle v\otimes w, Vu\rangle = \sum_a \langle v,\sqrt{\mu_a}u\rangle \langle w,a\rangle = \langle V^\dagger(v\otimes w),u\rangle.$$
With these, observe that
$$V^\dagger(I\otimes |a\rangle\!\langle a|) V \,u
= \mu_a u.$$
In other words, we can write the elements of the POVM as
$$\mu_a = V^\dagger(I\otimes |a\rangle\!\langle a|) V.$$
This is essentially the statement at hand: $V$ is an isometry that maps the states into a larger space, where a projective measurement in the basis $|a\rangle$ is performed.
The probabilities produced by the POVM then read
$$\langle \mu_a,\rho\rangle
= \langle I\otimes |a\rangle\!\langle a|, V\rho V^\dagger\rangle
\equiv \langle a| \operatorname{Tr}_1[V\rho V^\dagger]|a\rangle,$$
which is again directly interpreted as evolving $\rho$ through the isometry $V$, and then performing a projective measurement on the ancillary degree of freedom.
See e.g. Watrous' book, section 2.3, for more info.
Examples
You can find a few examples of POVMs in this other answer.
Consider here the somewhat trivial single-qubit two-outcome POVM with $\mu_1=\mu_2=\frac12 I$. This has elements that have rank 2, and following the procedure above we find dilations of the form
$$V = \frac1{\sqrt2}\sum_{a=1}^2 I\otimes |a\rangle.$$
In matrix form, this can be represented as
$$V = \frac1{\sqrt2}\begin{pmatrix}1&0 \\ 0&1 \\ 1& 0\\0&1\end{pmatrix}$$
I should point out that the way you represent $V$ (as any linear operator really) depends on the choice of basis. The matrix representation I'm using here would arguably be the one more naturally attached to $\frac12\sum_a |a\rangle\otimes I$ rather than $V$ itself, but as long as one knows how things are being represented there is no harm in using this representation.
Note also that the dilation isometry is not unique. It depends on our choice of basis in the ancillary space.
So, given an initial state $|\psi\rangle$, we get $V|\psi\rangle=|\psi\rangle\otimes|+\rangle$, and thus the outcome probabilities are $p(i) = \| (I\otimes \langle i|)(|\psi\rangle\otimes|+\rangle)\|^2=\frac12$, for $i=0,1$. Which is compatible with the POVM being such that $\langle\mu_i,\rho\rangle=\frac12$. Admittedly, this is a completely useless measurement which extracts no information from the state, I only use it to illustrate the procedure. Similar calculations can be performed in less trivial cases though, e.g. something like $\mu_1=I/2+\epsilon$ and $\mu_2=I-\mu_1$ for $\epsilon>0$ small enough.
