For countable $\tau$, I believe the answer is no. Joel Hamkins has commented on your Math.SE version of this question that from this we can get a negative answer for general $\tau$ by a forcing argument, giving a complete answer to your question. Joel, if you read this, it'd be great if you could give an explanation of your comment. I can edit my answer if need be. 

So assume that $\tau$ is a countable linear order, and that for some $n$ we have $\tau^n \cong \tau$. We show that in fact $\tau^2 \cong \tau$ (and so $\tau^m \cong \tau$ for every $m$). There are four possibilities: either $\tau$ has a top element (but no bottom), a bottom element (but no top), both a top and bottom element, or neither. For now let's assume we are in the last case, which is the simplest. The other cases are quite similar, and I'll describe how to handle them at the end.

The idea is to decompose $\tau$ in such a way that it is easy to see that $\tau^2 \cong \tau$. Forgetting $\tau$ for a moment, let's describe the kind of decomposition we are aiming for. First, decompose the set of rationals $\eta$ into countably many disjoint copies of itself. That is, write $\eta = \bigcup_{i \in \mathbb{N}} \eta_i$, where each $\eta_i \subseteq \eta$ is dense and for $i \neq j$ we have $\eta_i \cap \eta_j = \emptyset$. I'll refer to the elements of $\eta_i$ as $i$-points. One may think of $i$-points as being colored by the color $i$. 

Now, suppose for each $i \in \mathbb{N}$ we have some countable linear order $L_i$. We may form a new order $L$ by replacing (in $\eta$), for each $i \in \mathbb{N}$, each $i$-point with a copy of $L_i$. Somewhat informally, I'll write
\[
L = \bigcup_{i \in \mathbb{N}} \eta_i \times L_i.
\]
The claim is that $L^2 \cong L$. Indeed, if $\beta$ is *any* countable order, we have $\beta \times L \cong L$. For one may think of $\beta \times L$ as being formed in two steps: first, replace each point in $\beta$ with copy of $\eta$ to form $\beta \times \eta$, remembering our coloring of $\eta$. Then of course $\beta \times \eta$ is isomorphic to $\eta$, and moreover, for each $i$ the collection of $i$-points in $\beta \times \eta$ is dense. Thus there is an isomorphism between $\beta \times \eta$ and $\eta$ that sends $i$-points to $i$-points. (I am using here the fact that if $X$ and $Y$ are countable dense orders, each decomposed into countably many disjoint dense sets, $X = \bigcup_i X_i$, $Y = \bigcup_i Y_i$, then there is an isomorphism of $X$ and $Y$ that sends $X_i$ onto $Y_i$ for every $i$.) Since $\beta \times L$ is obtained by replacing each $i$-point in $\beta \times \eta$ with a copy of $L_i$, we see that $\beta \times L \cong L$, as claimed. Notice that the same argument works if some (or even all but finitely many) of the $L_i$ are empty. 

Our goal then is to decompose $\tau$ similarly as $\bigcup_i \eta_i \times L_i$ for some collection of countable order types $L_i$, for then by above we have $\tau^2 \cong \tau$. By assumption, we have $\tau \cong \tau^n$. Let us instead write $\tau \cong \beta \times \tau$, where $\beta = \tau^{n-1}$. For us, the form of $\beta$ is irrelevant except that it is countable and, like $\tau$, has neither a top nor bottom element. 

Now, the fact that $\tau \cong \beta \times \tau$, means that $\tau$ may be split up into $\beta$-many intervals, each of order type $\tau$ (where "$\beta$-many" is informal but hopefully clear). Each of these $\beta$-many copies of $\tau$ may in turn be split into $\beta$-many copies of $\tau$, and so on. This is as much to say that to every finite sequence $s = \langle x_1, x_2, \ldots, x_n \rangle$ of elements of $\beta$, we may associate an interval $I_s \cong \tau$. Namely, $I_s$ is the $x_n$-th copy of $\tau$ within the $x_{n-1}$-th copy of $\tau$ $\ldots$  within the $x_1$-st copy of $\tau$. We also have an $f_s: \tau \rightarrow I_s$ witnessing the isomorphism. 

Notice that in iteratively splitting $\tau$ into smaller and smaller copies of itself, we may choose the $I_s$ and $f_s$ in such a way that the maps $f_s$ commute, that is, so that if $s^{\frown}t$ is the sequence obtained by concatenating the sequences $s$ and $t$, then $f_{s^{\frown}t} = f_{s} \circ f_{t}$ (where, on the right, "$f_s$" really means "$f_s \upharpoonright I_t$"). This I think is really the natural thing to do, but to be clear let me say what I mean. At the first stage, for every $x \in \beta$ we have an interval $I_{\langle x \rangle}$ of order type $\tau$ and an isomorphism $f_{\langle x \rangle}: \tau \rightarrow I_x$. For every $y \in \beta$, let $I_{\langle y, x \rangle}$ be the image of $I_{\langle y \rangle}$ under $f_{\langle x \rangle}$. Then naturally we obtain the isomorphism $f_{\langle x, y \rangle} = f_{\langle x \rangle} \circ f_{\langle y \rangle}$ from $\tau$ onto $I_{\langle x, y \rangle}$. And so on for sequences of length longer than 2. Note that if $s$ extends $t$, then $I_s$ is a proper subinterval of $I_t$. 

Now, having associated to every finite sequence $s \in \beta^{<\omega}$ an interval $I_s$, we may associate to every infinite sequence $r \in \beta^{\omega}$ the interval $I_r = \bigcap_{n} I_{r \upharpoonright n}$ obtained by taking the natural nested intersection. The collection of $I_r$ for $r \in \beta^{\omega}$ is a covering of $\tau$ by nonintersecting intervals. Our construction guarantees that $I_r$ lies to the left of $I_s$ in $\tau$ if and only if $r < s$ in the lexicographical ordering of $\beta^{\omega}$. Note that for a fixed $r$, it may be that $I_r$ contains many points, a single point, or no points at all. Indeed, since $\tau$ itself is countable, $I_r$ will be empty for all but countably many $r$. The relevant observation for our purposes is that, if we view $I_r$ itself as a linear order, then for many densely $s \in \beta^{\omega}$ we have $I_r \cong I_s$. To see this, let us introduce an equivalence relation on the space $\beta^{\omega}$: say that $r \sim s$ iff $r$ and $s$ share a tail-sequence (not necessarily beginning at the same coordinate). That is:
\[
r \sim s \leftrightarrow \textrm{there exists $m, n$ such that for every $i$ we have $r(m+i) = s(n+i)$},
\]
where $r(k)$ means the $k$th entry in the sequence $r$, etc. This is clearly an equivalence relation. The claim is that if $r \sim s$, then $I_r \cong I_s$. For indeed, if $r \sim s$, then there are finite sequence $r', s' \in \beta^{<\omega}$ and an infinite sequence $x \in \beta^{\omega}$ such that $r = r'^{\frown}x$ and $s=s'^{\frown}x$. The intuitive reason that $I_r \cong I_s$ is that, in order to construct these intervals, we first move to $I_{r'}$ and $I_{s'}$, which are both just copies of $\tau$. Then, in each interval, we travel down the precisely corresponding nested sequence of intervals (represented by $x$) to obtain $I_r$ and $I_s$, which must therefore be isomorphic. More formally, we see that by the way we defined $f_{r'}$ and $f_{s'}$, we have $f_{s'} \circ f_{r'}^{-1}$ maps $I_r$ isomorphically onto $I_s$, establishing the claim. 

For $s \in \beta^{\omega}$, let $\bar{s}$ denote its equivalence class. It is easy to see that if $s_1, s_2 \in \bar{s}$ and $s_1 < s_2$ (where again $<$ is the lexicographical ordering on $\beta^{\omega}$), then there are $r, r', r'' \in \bar{s}$ such that $r < s_1 < r' < s_2 < r''$ (we use here that $\beta$ has neither a top nor bottom element). It follows that $\bar{s}$ has the order type of the rationals. By what we just showed, for a fixed $\bar{s}$, every $I_r$ with $r \in \bar{s}$ has the same order type. Since only countably many of the $I_r$ are nonempty, we may enumerate the (distinct) classes $\overline{s_1}, \overline{s_2}, \ldots$, such that if $I_r$ is nonempty, there is $i$ such that $r \in \overline{s_i}$ (this list may be finite). Let $L_i$ be the shared order type of every $I_r$, $r \in \overline{s_i}$. 

But now we have our desired decomposition. As already noted, every $\overline{s_i}$ has type $\eta$, and thus so does the subset $S \subseteq \beta^{\omega}$ given by $S = \bigcup_i \overline{s_i}$. We think of each $\overline{s}_i \subseteq S$ as our collection of $i$-points. But then by our choice of the $\overline{s_i}$, we have that $\tau$ may be formed by replacing each point $i$-point in $S$ by a copy of $L_i$. That is, $\tau = \bigcup_i \overline{s_i} \times L_i$. Then, by our argument given at the beginning, $\tau^2 \cong \tau$, as desired. 

This takes care of the case when $\tau$ has neither a top nor bottom element. The other three cases are similar, but with complicating details. I will sketch the variations that arise. Hopefully I have not overlooked anything. 

Suppose first that $\tau$ has a bottom element, but not a top one. In this case, the preliminary discussion given in paragraphs 3 and 4 is modified as follows. Instead of decomposing $\eta$ into countably many copies of itself, we decompose $1+\eta$ (that is, the order type of the rationals with a left endpoint added) as $1 + \eta_0 \, \, \cup \, \, \,\bigcup_{i>0} \eta_i$, where the $\eta_i$ are pairwise disjoint, and each is dense in $\eta$. This is the same as before, except that in our decomposition we have the distinguished set $1+\eta_0$ that includes the left endpoint, i.e. we color the left endpoint as a $0$-point. Suppose as before we are given a collection of countable orders $L_i$, where now we insist $L_0$ has a left endpoint. We may form $L$ by replacing every $i$-point with $L_i$, and write 
\begin{equation}
(1)     \, \, \, \, \, L = (1 + \eta_0) \times L_0 \, \, \cup \, \, \bigcup_{i>0} \eta_i \times L_i.      
\end{equation}

Note that since $L_0$ has a left endpoint, $L$ does as well. The claim is again that $L^2 \cong L$, or more generally that if $\beta$ is any countable order *also with a left endpoint*, then $\beta \times L \cong L$. The same argument as before goes through, the relevant difference being that whereas in the previous case we used the fact that $\beta \times \eta \cong \eta$ for any countable $\beta$, we now use the fact that since $\beta$ has a left endpoint, we have $\beta \times (1+\eta) \cong 1+\eta$ (so that $\beta \times (1+\eta)$ may be colored in the same way we colored $1+\eta$). 

Then, we seek to use $\tau^n \cong \tau$ to decompose $\tau$ in the form of (1). Again we write $\tau$ as $\beta \times \tau$, where $\beta = \tau^{n-1}$, and note that $\beta$ has a left endpoint, which we label as 0. We again associate to every $s \in \beta^{<\omega}$ an interval $I_s$ and isomorphism $f_s$ of $I_s$ with $\tau$. By taking intersections we obtain the intervals $I_s$ for $s \in \beta^{\omega}$. Again if $s \sim r$, then $I_s \cong I_r$. 

For each $s \in \beta^{\omega}$, the equivalence class $\bar{s}$ will again have the order type of the rationals, *except* for the equivalence class of the zero sequence $\mathbf{0} = \langle 0, 0, \ldots \rangle$. This class has order type $1 + \eta$, the left endpoint being $\mathbf{0}$ itself. Label this class as $\overline{s_0}$. Let $L_0$ be the shared order type of intervals associated to this class, and note that $L_0$ has a left endpoint. For $L_0$ is the order type of $I_{\mathbf{0}}$, which contains the left endpoint of $\tau$. We may enumerate the other classes corresponding to nonempty intervals as $\overline{s_1}, \overline{s_2}, \ldots$. Letting $L_i$ be the order type of the intervals associated to $\overline{s_i}$, we again see that $\tau$ may be written as $(\overline{s_0} \times L_0) \cup \bigcup_{i>0} \overline{s_i} \times L_i$. Since $\overline{s_0}$ has type $1+\eta$ and all other $\overline{s_i}$ have type $\eta$, this is a decomposition of the desired form (1). 

The case when $\tau$ has a top element but no bottom element is symmetric. Finally, suppose $\tau$ has both a top and bottom element. Briefly, the modifications in the argument are:  begin by decomposing $1+\eta+1$ as $1 + \eta_0 \, \cup \eta_1 + 1 \, \cup \, \bigcup_{i > 1} \eta_i$ (so the left endpoint is a 0-point, the right endpoint is a 1-point). Suppose that $L_0$ is a countable order with a left endpoint, $L_1$ is a countable order with a right endpoint, for $i>1$ the $L_i$ are arbitrary countable orders. Then if $L$ is obtained by replacing each $i$-point in $1 + \eta + 1$ with a copy of $L_i$, we have that $\beta \times L \cong L$ for any countable $\beta$ *also with a top and bottom point*. In particular, $L^2 = L$. Toward decomposing $\tau$ in the form of $L$, we write $\tau = \beta \times \tau$ where $\beta = \tau^{n-1}$ has a bottom point 0 and a top point 1. Our classes $\bar{s}$ for $s \in \beta^{\omega}$ will all have order type $\eta$, except for the class of $\overline{\mathbf{0}}$, which has type $1+\eta$, and the class $\overline{\mathbf{1}}$, which has type $\eta+1$. Letting $L_0$ be the order type associated to $\overline{\mathbf{0}}$, we see that $L_0$ has a left endpoint. Similarly, $L_1$ has a right endpoint, where $L_1$ is the type associated the $\overline{\mathbf{1}}$. Labeling the other classes corresponding to nonempty order intervals as $\overline{s_2}, \overline{s_3}, \ldots$ we similarly obtain the desired decomposition of $\tau$ in the form of $L$.