I think I can see an example using decomposition theory. I am mostly using the notation from the wikipedia article on the Whitehead manifold. I am not sure yet, whether it is an example for $n=3$ or $n=4$. Take $K_1=S^1\times D^2$ and $K_2=(S^1\times D^2) / W$ where $W=\bigcap_n W_i$ is the intersection of a sequence of nested solid tori such that each $W_{i+1}$ is embedded in $W_i$ as a Whitehead-link and we use the standard framing (although I am not sure whether this makes a difference for the argument).
Certainly $S^1\times D^2\times \mathbb{R}$ can be embedded into $\mathbb{R}^4$. Thus knowing that it is homeomorphic to $K_1\times \mathbb{R}$ would show that $K_1$ embeds into $\mathbb{R}^4$. I don't know whether it embeds into $\mathbb{R}^3$.
I believe that Bing shrinking can be used to show this and that I heard about this example in a lecture quite some time ago.
I looked it up in Theorem 7.5 in "Behrens, Stefan, Boldizsár Kalmár, Min Hoon Kim, Mark Powell, and Arunima Ray, eds. The disc embedding theorem. Oxford University Press, 2021.", which again gives
Andrews, J. J., and Leonard Rubin. "Some spaces whose product with E^1 is E^4." (1965): 675-677.
as a reference.
Maybe it is worth to sketch the main ideas of the construction/ of decomposition space theory here. Suppose we have a compact Hausdorff space $X$ together with a sequence of nested, compact subspaces $W_i$. and let $W_\infty =\bigcap_i W_i$.
First note that $X/W_\infty \to \lim_\leftarrow X/W_i$ is a homeomorphism, since it is a continuous bijection between compact Hausdorff spaces. Especially, if the diameter of $W_i$ goes to zero, we get that $W_\infty=pt$ and hence $X/W_\infty$ would be again $X$ in this case. Since the sets are nested, their diameters are monotonically decreasing, so we really just have to find a subsequence whose diameters go to zero.
But If that diameter does not go to zero, the trickery begins. Now suppose that we are in a situation, where for every $n\in \mathbb{N}$ and every $\varepsilon>0$, we can find a homeomorphism $X$ and a $N\ge n$ which is the identity outside of $W_n$ and such that the diameter of $\varphi(W_N)<\varepsilon$.
We then would like to find an isomorphism of inverse systems
$(X/W_i)_i\to (X/W'_i)_i$ where the diameters on the right hand side go to zero. A very basic example is the setting, where all $W_i$'s and $W'_i$'s are balls around 0 and the diameter of the $W_i$'s does not go to zero while the diameter of the $W'_i$'s goes to zero.
The first step is easy, choose $n=1$ and $\varepsilon=1$ and find $N=N_1$ and $\varphi$ such that $diam(\varphi(W_n))<1$. This suggests that we could pick $W'_k=\varphi(W_k)$ for $k=1,\ldots,N_1$. But what do we do afterwards? We have no control over the diameters of $\varphi(W_{N_1+1})$ and so on. So maybe we should have improved the situation there first. I am looking for a $\varphi_2$ and an $N_2$ such that $diam(\varphi(\varphi_2(W_{N_2})))<1/2$. Luckily $X$ is compact and so $\varphi$ is uniformly continuous, e.g. there exists an $\varepsilon$,such that if $diam(\varphi_2(W_{N_2}))<\varepsilon$, then $diam(\varphi(\varphi_2(W_{N_2})))<1/2$. And our magical way to find such homeomorphisms tells us that such an $\varphi_2$ exists. We can then define $W'_i = \varphi(\varphi_2(W_i))$ for $i=1,\ldots,N_2$ and our assumptions on the support tells us that this agrees with the previous definition. Continuing with this strategy, we get the desired map of inverse systems.
The big remaining question is how to construct these maps in specific cases.
Let us first have a look at a solid torus, e.g. $X=W_0=S^1\times D^2$ and each $W_{i+1}$ is obtained from $W_i$ using the Whitehead-link. Sadly in this case $X/W_\infty$ cannot be homeomorphic to a solid torus, as removing the point $[W_\infty]$ gives a space with fundamental group at infinity, very much like the Whitehead-manifold.
Now there are some issues why the situation here is more difficult. The first is that $S^1\times D^2\times \mathbb{R}$ is not compact, so all arguments break down. Thatfor I would like to alter the question to show that $(S^1\times D^2)/W_\infty\times S^1$ is homeomorphic to $(S^1\times D^2)\times S^1$ and then the corresponding covers are homeomorphic and this is what we are looking for.
The second issue is that we are not just collapsing one compact set, but now our equivalence relation is given by collapsing an one parameter family of compact sets each to a different point, e.g. we have a sequence of nested, compact equivalence relations $\sim_n\subset X^2$. Actually the same ideas work if we replace each occurence of $W_n$ by each of the equivalence classes of $\sim_n$.
The naive idea that things are only knotted because we are missing one extra dimension does not work. moving in the new $S^1$-direction we have knotted tori over each point in $S^1$. We need to do something trickier.
Define a continuous map $W_1\to \mathbb{R}$ as follows. Note that the winding number of the embedding $W_1\to W = S^1\times D^2$ in the Whitehead-link is zero. We thus can project to $S^1$, lift that map to $\mathbb{R}$, and then multiply with $1/n$ for large $n$. This defines a continuous map $W_1\to \mathbb{R}$. Extend it to the entire $W_0$.
Then define a self-homeomorphism of $S^1\times D^2\times S^1, (w,t)\mapsto (w,t+f(w))$. This has the effect that each equivalence class, which was previously concentrated at a specific $t\in S^1$ now slightly spirals upwards, e.g. it still goes once around the first $S^1$ coordinate, links with some other stuff and then goes back. But now in the second $S^1$-coordinate it also goes up by roughly $1/n$ before it goes back.
The final trick is to use the homeomorphism of $S^1\times S^1$, which sends $(x,y)$ to $(ny+x,y)$. Then it does not even go around the first
$S^1$ anymore and you can make the diameter arbitrarily small.
I do now that this is just a sketch, but maybe I could show the ideas. More detailed (and more formal) proofs should be given in the references.
