Yes, decimating an $m$-sequence of length $L=2^k-1$ generated by a primitive feedback polynomial by taking every $i$th bit yields another $m$-sequence whenever $\gcd(i,L)=1$. So in the Mersenne case when $L$ is a prime number, we get this for all decimation exponents $i, 0<i<L$.
My answer uses a few facts about finite fields: their construction as well as trace function. If you have not heard of them, then this may be difficult to follow. As you seem to know what a primitive polynomial means in the context of finite fields, you probably won't have too many difficulties. Do ask, if you need clarifications.
If $m(x)\in\Bbb{F}_2[x]$ is a primitive polynomial of degree $k$, and $\alpha$ is its zero in the extension field $K=\Bbb{F}_{2^k}$, then $\ell=L$ is the lowest positive exponent with the property that $\alpha^\ell=1$. This is the defining property of primitive polynomials (in this context), and $\alpha$ is called a primitive element of $K$ as it generater the multiplicative group $K^*$.
If $tr:K\to\Bbb{F}_2$ is the trace function:
$$
tr(x)=x+x^2+x^4+x^8+\cdots+x^{2^{k-1}}=\sum_{j=0}^{k-1}x^{2^j},
$$
then the $m$-sequences $(s[i])_{0\le i<L}$ generated by the feedback polynomial $m(x)$ all have the form
$$
s[t]=tr(\alpha^{t+\tau}), t=0,1,\ldots,L-1.
$$
Here $\tau$ gives the phase shift of the sequence, and the sum $t+\tau$ can be computed modulo $L$ as $\alpha^L=1$. If
$$
m(x)=\sum_{j=0}^km_jx^j=1+m_1x+m_2x^2+\cdots+m_{k-1}x^{k-1}+x^k,
$$
then for all $t$ we first get the relation
$$
0=\sum_{j=0}^km_j\alpha^{t+j}\tag{1}
$$
by multiplying the equation $m(\alpha)=0$ by $\alpha^t$. And then, because the trace respects sums
$$
0=\sum_{j=0}^k m_j s[t+j].\tag{2}
$$
In other words, the sequence $(s[t])_{0\le t<L}$ satisfies the recurrence relation corresponding to the feedback polynomial $m(x)$. Caveat: IIRC some authors use the reciprocal polynomial $\tilde{m}(x)=x^km(1/x)$ here. This is something I always need to check when reading any source :-)
The idea here is that the choice of the phase shift $\tau$ ($L$ possible values) gives us the freedom to initialize the linear feedback shift register with any combination of bits. Or, any combination of choices for the bits $s[0], s[1],\ldots, s[k-1]$ such that at least one of them is $\neq0$ can be obtained with a careful choice of $\tau$. This is the magic of $m$-sequences.
On with the decimations. Let us select a decimation exponent $i$ such that $\gcd(i,L)=1$. We are looking at the decimated sequence
$$s'[t]:=s[ti],\tag{3}$$
where, again, $ti$ is computed modulo $L$. So
$$s'[t]=tr(\alpha^{ti+\tau}).$$
From the basic properties of cyclic groups it follows that $\beta=\alpha^i$ is another primitive element. Consequently we can find an exponent $\tau'$ such that
$\beta^{\tau'}=\alpha^\tau$. Hence
$$
s'[t]=tr(\alpha^{ti}\cdot\alpha^\tau)=tr(\beta^{i+\tau'}).\tag{4}
$$
Tracing our earlier steps backwards we see that if $m'(x)$ is the minimal polynomial of $\beta$, then the decimated sequence satisfies the linear feedback recurrence corresponding to the primitive polynomial $m'(x)$. This proves the main claim.
To give us a way of verifying the above theory let's consider the case
$$m(x)=x^5+x^2+1.$$
By the so called Freshman's dream (of characteristic two fields) we see that
$$
m(\alpha^2)=\alpha^{10}+\alpha^4+1=(\alpha^5+\alpha^2+1)^2=m(\alpha)^2=0.
$$
In other words, with decimation exponent $i=2$ we see that the decimated $m$-sequence satisfies the same recurrence relation as the original one — something you observed yourself already. This happens always. Whenever we decimate such an $m$-sequence with an exponent that is a power of two, we get back the same
sequence (possibly with a different shift).
With the decimation exponent $i=3$ the situation is different in the sense that $\beta=\alpha^3$ has a different minimal polynomial. I shamelessly refer you to an earlier answer of mine, where I try to explain how we can see that in this very instance the minimal polynomial of
$\beta$ is
$$m'(x)=1+x^2+x^3+x^4+x^5.$$
As a check I invite you to verify that the $m$-sequence you get, starting from your given sequence $S$ and using the decimation exponent $i=3$, satisfies the recurrence relation corresponding to the feedback polynomial $m'(x)$.
