"Phantom" non-equivalences of spectra? I would like an example of the following situation, or a proof that no such example exists.
$\textbf{Situation}$: Two connective (EDIT: I'm fine with dropping this condition) spectra $X$ and $Y$ such that the spaces $\Omega^\infty\Sigma^nX$ and  $\Omega^\infty\Sigma^nY$ are equivalent (as spaces) for all $n$, but as spectra, $X$ and $Y$ are $\textit{not}$ equivalent. It was pointed out to me by Maxime that another way to phrase things is that $X$ and $Y$ are two (grouplike) $E_\infty$ spaces that are equivalent as $E_n$-spaces for all finite $n$ but not as $E_\infty$-spaces.
I think (as Charles pointed out in the comments, this is wrong) it would be sufficient to find an example of the following: An infinite loop space $A$ (with terms $A_0=\Omega A_1,A_1,...$ and structure maps $a_i:A_i\rightarrow \Omega A_{i+1}$ such that $a_0$ is the identity) and an equivalence $f:A_0\rightarrow A_0$ (of spaces) such that for any (infinite loop) equivalence $\phi:A\rightarrow A$ with 0th component $\phi_0:A_0\rightarrow A_0$, the composite $\phi_0\circ f$ is not homotopic to the component of any infinite loop map. Because then we can consider the infinite loop space $B$ with the same terms $B_i=A_i$ and the same structure maps $b_i=a_i$ for $i>0$ but with the $b_0$ changed to $f$. The condition on $f$ implies that there is no way to fill in the 0th component of any attempted equivalence between $A$ and $B$.
The problem is that due to my insanely low supply of brain cells the only infinite loop spaces I can really come to grips with are Eilenberg-Maclane, and among Eilenberg-Maclane spectra such an example can't exist since those determined by their homotopy groups...
 A: I am pretty sure that it will be rather delicate to find an example like you want. Here are some thoughts:
Let $X_n$ denote $\Omega^{\infty} \Sigma^n X$.  One is assuming an equivalence $X_n \simeq Y_n$ for all $n$, and asking about an equvalence $X \simeq Y$.
Let $\mathcal S_n= \{f: X_n \simeq Y_n \}\subset [X_n,Y_n]$. Note that looping gives set maps $\mathcal S_{n+1} \rightarrow \mathcal S_n$.  It is not hard to check that there is an equivalence $X \simeq Y$ if and only if the inverse limit of the $\mathcal S_n$ is nonempty.  So one would have an equivalence if this were an inverse limit of finite sets, or profinite sets, etc., and with many spectra $X$ that come to mind, this would be the case.
A: Here's a connective example. It is also an example of Maxime's variant question in the comments (regarding $\tau_{\leq m}$ truncations). And thanks to Maxime for looking this argument over before I posted it (but any errors are mine, etc etc).
It's just a tweaked version of Tyler's example and the arguments are very similar but I repeated them for the sake of self-contained-ness. Define $B=\prod_{n=2}^\infty H\mathbb{Z}/2$ and $C=\prod_{n=2}^\infty\Sigma^nH\mathbb{Z}/2$. Consider the following two maps $h,j:B\rightarrow C$. The map $h$ is the infinite product of the maps $Sq^n:H\mathbb{Z}/2\rightarrow \Sigma^n H\mathbb{Z}/2$. The map $j$ is the precomposite of $h$ with the shift-down self-map of $B$. Said differently we can write $B=H\mathbb{Z}/2\oplus \prod_{n=3}^\infty H\mathbb{Z}/2$ and $C=0\oplus\prod_{n=2}^\infty\Sigma^nH\mathbb{Z}/2$ and $j$ sends the first summand of $B$ to zero, and on the second summand is the infinite product of $Sq^{n-1}$. Let $Z$ and $W$ be the fibers of $h$ and $j$. Note that $Z\simeq\prod_{n=2}^\infty\text{cofib}(Sq^n)$ and $W\simeq Z\oplus H\mathbb{Z}/2$.
The truncations
$\tau_{\leq m}Z$ and $\tau_{\leq m}W$ are given by truncations of the fibers of the maps
$$\prod_{n=2}^\infty H\mathbb{Z}/2\xrightarrow{\tau_{\leq m}h,\tau_{\leq m}j}\prod_{n=2}^m \Sigma^nH\mathbb{Z}/2.$$
But note that $\tau_{\leq m}h$ and $\tau_{\leq m}j$ are equivalent since they are the same finite list of $Sq^k$'s together with zero on infinitely many factors of the domain $B$. Swindled! Hence $\tau_{\leq m}Z\simeq\tau_{\leq m}W$ for all $m$.
Let's show that $Z$ and $W$ are not equivalent. Recall that $W$ is equivalent to $Z\oplus H\mathbb{Z}/2$. So it suffices to show that $Z$ has no $H\mathbb{Z}/2$ summand. For a contradiction, suppose $Z\simeq V\oplus H\mathbb{Z}/2$. That $H\mathbb{Z}/2$ summand has to map nontrivially to $B$, for if it mapped trivially then $C$ would have a $\Sigma H\mathbb{Z}/2$ summand, but $\pi_1C=0$. Now any nontrival map from $H\mathbb{Z}/2$ to $B$ maps via the nontrivial self map of $H\mathbb{Z}/2$ to a nonzero number of the defining $H\mathbb{Z}/2$-factors of $B$ (after all $B$ is a $\mathbb{Z}/2$-vector space). The composite of such a map with $h$ cannot be null since $h$ is non-null on each factor, and the factors do not interact in the target (they map to diferent factors). So there is no $H\mathbb{Z}/2$ summand in the fiber of $h$ (which is $Z$).
Lastly, let's show that for all $m$ we have $\Omega^\infty\Sigma^mZ\simeq\Omega^\infty\Sigma^mW$ as spaces. Now consider $B$ and $C$ in their standard $\Omega$-spectrum presentations by Eilenberg-Maclane spaces. We will use the fact that the fiber of a map of $\Omega$-spectra is computed levelwise and is again an $\Omega$-spectrum. So to compute that $k$-th space in an $\Omega$-spectrum presentation of $\Sigma^{-1}Z$ and $\Sigma^{-1}W$, we need to compute the fiber of the map $h_k, j_k:\prod_{n=2}^\infty K(\mathbb{Z}/2,k)\rightarrow\prod_{n=2}^\infty K(\mathbb{Z}/2,k+n)$ induces by $h$ and $j$. But for $n>k$, the map $Sq^n:K(\mathbb{Z}/2,k)\rightarrow K(\mathbb{Z}/2,k+n)$ is null, so again, $h_k$ and $j_k$ are the same finite lists of unstable $Sq^i$'s plus a bunch of nullhomotopic maps, so they are equivalent by another swindle.
A: For this we can use a swindle-type technique.
Let $B = \bigoplus_{n=2}^\infty H\Bbb Z/2$, and $A = \bigoplus_{n=2}^\infty \Sigma^{-n} H\Bbb Z/2$. We can construct maps $A \to B$ by specifying their effect on each summand of $A$.

*

*The first map, $f: A \to B$, is the sum of the composites $$\Sigma^{-n} H\Bbb Z/2 \xrightarrow{Sq^n} H\Bbb Z/2 \to \bigoplus_{n=2}^\infty H\Bbb Z/2,$$ where the second map includes as the $n$'th summand.

*The second map, $g: A \to B$, is the same, except the we map the $n$'th summand of $A$ to the $(n+1)$'st summand of $B$. (In particular, the summand with $n=2$ is completely missed by $g$.)

Define $X = cofib(f)$ and $Y = cofib(g)$. Since these sums start at $n=2$, $B$ is the connective cover of both $X$ and $Y$ by the long exact sequence in homotopy. I claim that $X$ and $Y$ have equivalent connective covers $\tau_{\geq m} X$ and $\tau_{\geq m} Y$ for all $m < 0$, and that they are inequivalent spectra. This suffices to give an example for your question, since taking the associated infinite loop spaces factors through some connective cover.

First let's show that they have equivalent covers. Note that $\tau_{\geq m+1} X$ is the cofiber of a map
$$
\bigoplus_{2 \leq n \leq -m} \Sigma^{-n} H\Bbb Z/2 \xrightarrow{\tau_{\geq m} f} \bigoplus_{n=2}^\infty H\Bbb Z/2
$$
and similarly for $\tau_{\geq m+1} Y$. However, the maps $\tau_{\geq m} f$ and $\tau_{\geq m} g$ are equivalent maps in the homotopy category: both are the the same finite list of Steenrod squares summed with countably many extra summands of $H\Bbb Z/2$ on the target. Thus, these connective covers are equivalent.

Next, let's show that $X$ and $Y$ are not equivalent. To see this, first note that $Y$ has $H\Bbb Z/2$ as a summand: the map $g: A \to B$ completely misses the summand $H\Bbb Z/2$ and allows us to split it off. However, the cofiber sequence defining $X$ gives an exact sequence
$$
[X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \to [A, H\Bbb Z/2]
$$
which is more explicitly
$$
[X,H\Bbb Z/2] \to \prod_{n=2}^\infty H^0(H\Bbb Z/2) \xrightarrow{\prod Sq^n} \prod_{n=2}^\infty H^n(H\Bbb Z/2).
$$
This second map is injective, and so by exactness the restriction
$$
[X,H\Bbb Z/2] \to [B, H\Bbb Z/2] \cong Hom(\pi_0 X, \Bbb Z/2)
$$
must be trivial. Therefore, $X$ can't have $H\Bbb Z/2$ as a summand and thus can't be equivalent to $Y$.
