Yes, there is a proof which is long but one might consider to be "easy" after digesting it.
Let's first reduce show that the 2nd question (of triviality of connected finite etale covers) for the moduli stack $M_1$ is equivalent to its counterpart for the 1st one"Deligne-Rapoport" compactification $\overline{M}_1$ (a regular proper DM stack), to get since the method of the harder direction will be used in our argument for the case of $M_1$. The easier direction is that out if the case of $\overline{M}_1$ is known then we can settle the way. If case of $M_1$. It suffices to show that if a normal noetherian DM (or Artin) stack $X$ has a dense open substack $U$ with no nontrivial connected finite etale covers cover then the same holds for $X$. If It suffices to show more generally that if $U$ is a dense open substack of $X$ and $X' \rightarrow X$ is a finite etale cover then we can pick a section and $s:U \rightarrow X'|_U$ is a section over $U$, and this U$ then $s$ uniquely extends over $X$. Indeed the The uniqueness allows us to work over a smooth scheme chart, so we're reduced to the well-known case when $X$ is a scheme , where we (though one can feel more comfortable about also adapt to the case of stacks the proof in the scheme case via chasing connected componentscomponents).OK, so that's all
The more interesting direction is the converse: the case of $M_1$ implies the case of $\overline{M}_1$. For this we need just have to say about show that for any connected finite etale cover $T$ of $M_1$ the 2nd question connected finite normalization $\overline{T} \rightarrow \overline{M}_1$ (which is flat since I am assuming you meant for your cubic curves to have $\overline{M}_1$ is a section given through regular DM stack of dimension 2) is etale around the smooth locusclosed complement $\infty$ of $M_1$ in $\overline{M}_1$. Since $\overline{M}_1$ is regular and $\infty$ is a relative Cartier divisor that is regular with generic characteristic 0, so it follows from the relative Abyhankar Lemma (Exp. XIII, SGA1 for schemes, easily adapts to DM stacks by Deligne-Rapoport there usual etale-localization stuff) that $\overline{T}$ is a unique "relatively tamely ramified" along $\infty$. Thus, the etale-local structure near $\infty$ is given by an $e$th-root extraction of generalized elliptic curve, etc.).
As for a local generator of the first questionideal of the connected substack $\infty$ with $e$ a unit along $\infty$. But (as in Saito's argument mentioned by Minhyong) ${\rm{Spec}} \mathbf{Z}$ supports points of every possible prime residue characteristic, so $e$ has no prime factors and hence $e = 1$, so $\overline{T}$ is etale over $\overline{M}_1$ as desired.
Now we will directly attack the case of $M_1$ (though one can also solve it by the case of $\overline{M}_1$ directly in a more illuminating "topological" manner over $\mathbf{C}$ after some preliminaries with the triviality of $\pi_1({\rm{Spec}}(\mathbf{Z}))$ to handle geometric connectivity of connected components, but that gets caught up in "foundational" issues related to analytification of stacks). We shall initially working work with the open substack $M$ of elliptic curves with $j \ne 0, 1728$ on fibers; i.e., the stack of elliptic curves whose automorphism scheme is the constant group $\langle \pm 1 \rangle$ (exercise: equivalent to impose this condition on automorphism groups of geometric fibers, since that constant group has no nontrivial automorphisms); one might say this is the moduli stack of elliptic curves with "no extra automorphisms". I claim that $M$ has exactly one nontrivial connected finite etale cover, a scheme cover of degree 2, and we'll use this to bootstrap to get a handle on the entire moduli stack.
OK, now we will use Ihara's result noted by Minhyong to prove
Let's grant that $Y$ has trivial etale fundamental group. Observe that inside of the projective line over $\mathbf{Z}$, the open subscheme $Y$ (complement to the sections $0, 1728, \infty$) is very similar to the open subscheme $U$ complementary to the union of the disjoint sections $0, 1, \infty$. More precisely\pi_1(Y) = 1$, $j \mapsto j/1728$ defines an isomorphism between $Y$ and $U$ over $\mathbf{Z}[1/6]$, and this extends in both direction across the unique codimension-1 points in characteristics 2 and 3 also (via the valuative criterion for properness applied to the projective line over $\mathbf{Z}$). [ERROR: these maps carry the generic points in characteristics 2 and 3 to the closed points $0$ or $\infty$ on the other side. Probably JSE knows see how to circumvent this mistake.] Thus, it is an isomorphism away from some finite set of codimension-2 closed points supported in characteristics 2 and 3conclude. Thus, by Zariski-Nagata purity of the branch locus (these being 2-dimensional regular connected schemes), Then at the etale fundamental groups of $Y$ and end we will prove $U$ coincide\pi_1(Y) = 1$. Thus, the triviality result of Ihara for $U$ mentioned by Minhyong implies the same for $Y$.
Now consider Consider a connected finite etale cover $q:M' \rightarrow M$ of degree $> 1$. I claim it is isomorphic to $f$. Consider the pullback $Y' \rightarrow Y$ of $q$ along $f:Y \rightarrow M$. We just showed that Since $Y$ has trivial fundamental group, so this pullback splits as a disjoint union of copies of $Y$. Choosing such a component of the pullback defines a morphism $s:Y \rightarrow M'$ over $M$. But $f$ and $q$ are finite etale maps, so $s$ is also a finite etale map. But $M'$ is connected, so the open and closed image of $s$ is full; i.e., $s$ identifies $Y$ as a finite etale cover of $M'$, so we conclude that $M'$ is sandwiched inside the degree-2 finite etale cover $f$. But $q$ has degree $> 1$, so it follows that $q = f$ as desired.
OK, now we can finally solve the original problem . Let for $M_1$ denote (conditional on the triviality of $\pi_1(Y)$). The moduli stack of elliptic curves, so $M_1$ is regular and connected with $M$ a dense open substack that we have just seen has exactly one nontrivial connected finite etale cover. Let $M'_1 \rightarrow M_1$ be a connected finite etale cover with degree $> 1$. We seek a contradiction. The restriction over $M$ is a finite etale cover $M' \rightarrow M$ of degree $> 1$, and since $M'$ is open in the connected regular stack $M'_1$ it must also be connected. Thus, $M'$ is $M$-isomorphic to $Y$ (over $M$ via $f$). Hence, the elliptic curve $E$ over $Y$ extends to an elliptic curve $E'_1$ over $M'_1$ (namely, the pullback of the universal elliptic curve over $M_1$!).
The integral structure has done its job, and now to get the contradiction we consider a connected etale scheme neighborhood $(S,s)$ of a point $\xi$ with $j=0$ (or $j=1728$) on the DM stack $M'_1$ considered over $\mathbf{Q}$. We extend $S$ to a smooth connected complete curve $\overline{S}$ (with constant field that might be larger than ${\mathbf{Q}}$, but that won't matter for the ramification considerations we are about to undertake). Clearly $\overline{S}$ is a finite flat cover of the projective $j$-line over $\mathbf{Q}$ and at the point $s$ over $j=0$ (or $j=1728$) it has ramification degree 4 or 6 (I can't remember which is which) because of etaleness over $M_1$ and the fact that $M_1$ over the $j$-line has ramification over $j=0$ (or $j=1728$) equal to 4 or 6 (due to deformation theory considerations). By design, there is an elliptic curve over the open curve $S$ (namely, the pullback of the elliptic curve $E'_1$ over $M'_1$ that extends the elliptic curve $E$ over $Y$) whose discriminant in the function field of $\overline{S}$ is $(j(j-1728))^{-1}$ (well-defined up to 12th powers of nonzero elements, of course). But the "good reduction" at $s \in S$ forces the discriminant of any model over the function field of $S$ to have valuation at $s$ that is a multiple of 12, whereas for $(j(j-1728))^{-1}$ this valuation is $-4$ or $-6$ (since $S$ at $s$ has ramification over the $j$-line equal to 4 or 6). This is a contradiction, so $M_1$ has no nontrivial connected finite etale cover, assuming $\pi_1(Y) = 1$.
Finally, we prove $\pi_1(Y) = 1$. Note that $Y$ is the open complement in $\mathbf{P}^1_{\mathbf{Z}}$ of the union of the sections $\infty$, $j=0$, and $j=n$ with $n = 1728$. We will now work with any nonzero integer $n$. Since $\infty$ is disjoint from the others, by Saito's argument with the relative Abhyankar's Lemma as explained above, we see that any finite etale cover of $Y$ has normalization over that projective $j$-line over $\mathbf{Z}$ that is etale over $\infty$. Hence, it suffices to show that the open complement of $j(j-n)=0$ in $\mathbf{P}^1_{\mathbf{Z}}$ has trivial $\pi_1$. Making the change of coordinates $t = 1/j$ (which moves $j = 0$ out to $\infty$), this open complement is identified with the open complement $U_n$ in the affine $t$-line $\mathbf{A}^1_{\mathbf{Z}}$ of the locus $nt=1$. So it is enough to prove that $\pi_1(U_n) = 1$. Equivalently, we claim that $U_n$ has no nontrivial Galois connected finite etale covers. This will rest on two special facts about $\mathbf{Z}$: the triviality of $\pi_1({\rm{Spec}}(\mathbf{Z}))$ and the smallness of the group of roots of unity in $\mathbf{Z}$. Beware that $U_n(\mathbf{Z})$ is empty when $n \not\in \mathbf{Z}^{\times}$.
Let $h:V \rightarrow U_n$ be a Galois connected finite etale cover with degree $> 1$, so over $\mathbf{Q}$ we get a nontrivial connected finite etale cover $V'$ of the $\mathbf{Q}$-fiber $U'_n$ of $U_n$. We first claim that $V'$ must be geometrically connected over $\mathbf{Q}$. Since we're in characteristic 0, this amounts to the condition that $V'$ has constant field $\mathbf{Q}$. If we let the number field $K$ be its constant field then by normality of $V$ it follows that $h$ factors through $(U_n)_{O_K}$, with $V \rightarrow (U_n)_{O_K}$ necessarily surjective. This forces $(U_n)_{O_K}$ to be etale over $U_n$ (since $h$ is a finite etale cover), so since $U_n$ is fpqc over ${\rm{Spec}}(\mathbf{Z})$ it follows that ${\rm{Spec}}(O_K)$ is etale over ${\rm{Spec}}(\mathbf{Z})$. This forces $K = \mathbf{Q}$, as desired.
By the coordinate change $x = t/n$ we identify $U'_n$ with ${\rm{GL}}_1$, so $V'$ is a geometrically connected cover of ${\rm{GL}}_1$ over $\mathbf{Q}$. Thus, for $d = {\rm{deg}}(h) > 1$ we see that over an algebraically closed extension $k$ of $\mathbf{Q}$ the map $h_k$ is identified with the endomorphism $x^d$ of ${\rm{GL}}_1$. That is, the map $h'$ induced by $h$ between $\mathbf{Q}$-fibers is a "$\mathbf{Q}$-form" of the $\mu_d$-torsor ${\rm{GL}}_1$ over ${\rm{GL}}_1 = U'_n$. The set of isomorphism classes of such forms is given by $${\rm{H}}^1({\rm{GL}}_1,\mu_d) = (\mathbf{Q}^{\times}/({\mathbf{Q}}^{\times})^d) \times x^{\mathbf{Z}/d\mathbf{Z}}$$(since ${\rm{GL}}_1$ has trivial Pic and has unit group $\mathbf{Q}^{\times} x^{\mathbf{Z}}$).Explicitly, for $q \in \mathbf{Q}^{\times}$ and $j \in \mathbf{Z}$ the finite etale cover of ${\rm{GL}}_1 = U'_n$ associated to the class of $(q,j \bmod d)$ is given by the covering equation $y^d = q x^j$. As a covering of ${\rm{GL}}_1$ with coordinate $x$, this has geometric covering group $\mu_d(\overline{\mathbf{Q}})$ via scaling on $y$, so by inspection this geometric covering group action is not defined over $\mathbf{Q}$ (i.e., the automorphism group scheme for the covering is not a constant group over $\mathbf{Q}$, or in other words not all of these geometric automorphisms are defined over $\mathbf{Q}$) except when $d = 2$. Ah, but recall that we arranged for $h$ to be a Galois covering, so in our setting with the covering $h'$ the geometric covering group must be defined entirely over $\mathbf{Q}$ (as a constant group). In particular, $h$ must have degree $d = 2$. Also, the geometric connectedness of the covering forces ${\rm{gcd}}(j,d) = 1$.
To summarize, we have proved that $V \rightarrow U_n$ viewed over $\mathbf{Q}$ is given by $y^2 = q(t/n)$ for some $q \in \mathbf{Q}^{\times}$. By changing $y$ by a $\mathbf{Q}^{\times}$-scaling (as we may certainly do), we can change $q$ by any square multiple we wish, so we can arrange that $q/n$ is equal to a squarefree integer $r$. Then $V$ is identified with the normalization of $\mathbf{Z}[t][1/(nt-1)]$ in the $(nt-1)$-localization of $\mathbf{Z}[y,t]/(y^2 - rt)$. Using that $r$ is a squarefree integer, we claim that $\mathbf{Z}[y,t]/(y^2 - rt)$ is normal (in contrast with the situation for $\mathbf{Z}[y]/(y^2 - r)$ when $r$ is odd!). This is clear after inverting 2, so by Serre's homological criterion the only issue is to check the normality at the generic points in characteristic 2, which is to say that the maximal ideal of the local ring at these points is principal. If $r$ is even (so it is twice an odd integer) then $y$ lies in such primes, so $(2,y)$ is the only such prime and $t$ isn't in this prime. Hence, in such cases $y$ is a local generator (as the equation $rt = y^2$ with $t$ a local unit and $r$ twice an odd integer makes $2$ a local unit multiple of $y^2$). If $r$ is odd then $(2)$ is itself prime because the reduction of $y^2 - rt$ modulo 0 is the element $y^2 - t \in \mathbf{F}_2[t]$ that is irreducible.
We conclude that
$$V = {\rm{Spec}}(\mathbf{Z}[y,t]/(y^2 - rt))_{nt-1}$$over $U_n = {\rm{Spec}}(\mathbf{Z}[t])_{nt-1}$. Ah, but this is not etale over $U_n$, since passing to characteristic 2 turns this into a dense open piece of a purely inseparable quadratic cover in characteristic 2. Contradiction, so $\pi_1(U_n) = 1$. QED
