= Solution
Use the fact that <infinite cyclic subgroups of hyperbolic groups are undistorted>: if $z$ has infinite order in a <hyperbolic group>, then its <stable word length>
$$
\tau(z)=\lim_{m\to\infty}\frac{|z^m|}{m}
$$
is strictly positive. The limit exists by subadditivity and the <Fekete lemma>. It is invariant under <conjugation>, because conjugating $z^m$ changes its length by at most twice the conjugator's length. It also satisfies $\tau(z^k)=|k|\tau(z)$.
In the given <Baumslag-Solitar group>, $b$ has infinite order. To verify this without presuming the presentation's normal form, let $a$ act on the real line by $t\mapsto3t$ and $b$ by $t\mapsto t+1$. These affine bijections satisfy $aba^{-1}=b^3$, and every nonzero power of $b$ is a nonidentity translation. Thus the element $b$ in the presented group cannot have finite order.
If an injective <group homomorphism> to $G$ existed, its images $z$ of $b$ and $u$ of $a$ would satisfy $uzu^{-1}=z^3$, with $z$ still of infinite order. Consequently
$$
\tau(z)=\tau(uzu^{-1})=\tau(z^3)=3\tau(z),
$$
forcing $\tau(z)=0$, a contradiction. Therefore
$$
\boxed{\text{There is no injective homomorphism }\operatorname{BS}(1,3)\to G.}
$$
Equivalently, the relation gives $a^kba^{-k}=b^{3^k}$, so powers of $z$ with exponent $3^k$ would have length at most $2k|u|+|z|$. This contradicts the linear lower bound for an undistorted infinite cyclic subgroup. This alternative also makes the exponential compression obstruction explicit.
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