The answer is yes. And more generally, one has the following result:
Lemma: for any elliptic curve $X$, the degree of injective morphisms $X\to \mathbb{P}^2_{\mathbb{C}}$ (or $\mathbb{C}P^2$ if you want) is not bounded.
Proof: The curve $X$ is isomorphic to a plane cubic of equation $X^3+Y^3+Z^3=\lambda XYZ$ for some $\lambda$ (Hessian form). Then, the curve has nine inflexion points. These are points $[0:1:\omega]$, $[1:0:\omega]$, $[1:\omega:0]$ with $\omega^3=1$. Choose three of them, $p_1,p_2,p_3$ such that the three corresponding tangent lines $L_1,L_2,L_3$ intersect at three points $L_1\cap L_2$, $L_1\cap L_3$, $L_2\cap L_3$, different from $p_1,p_2,p_3$.
We change the coordinates and assume that $L_1,L_2,L_3$ are the lines $x=0$, $y=0$ and $z=0$. The points $p_1,p_2,p_3$ become points of the form $[0:1:a_1]$, $[a_2:0:1]$, $[1:a_3:0]$ with $a_1a_2a_3\not=0$.
Take now the birational map of $\mathbb{P}^2$ given locally by $(x,y)-->(x,x^ny)$, and globally by $[x:y:z]-->[xz^n:x^ny:z^{n+1}]$, for $n\ge 2$. Its base-points are $[0:1:0]$ and $[1:0:0]$, which do not belong to the curve. The curves contracted are $z=0$ and $x=0$, and $y=0$ is fixed. Outside of the triangle $xyz=0$, the map is an automorphism. In consequence, the image of the curve has degree $3n$ and only cuspidal singularities. This yields an injective map from $X$ to $\mathbb{P}^2$.
