Let $\varphi\in{\rm End}(G)$ s.t. $\exists n\ge 0$, $\ker(\varphi^n)=G$. If $\ker\varphi$ or $[G:{\rm Im}\varphi]$ is finite, then $G$ is finite

Assume that $\varphi$ has finite kernel. Consider the map $\varphi:\ker(\varphi^{k+1}) \rightarrow \ker(\varphi^k)$. It has finite kernel. So if it has finite range, it has finite domain. By induction, if $\ker(\varphi)$ is finite, so is every $\ker(\varphi^n)$ and in particular $G$.

If the image of $\varphi$ has finite index, then $\varphi^k(\varphi(G))$ has finite index in $\varphi^{k}(G)$ for all $k$. It follows that $1=\varphi^n(G)$ has finite index in $G$ so that $G$ is finite.

For the case where $\ker\varphi$ is finite:

Note that if $\ker(\varphi)\leq K\leq H$, then $[H:K]=[\varphi(H):\varphi(K)]$, by the isomorphism theorems.

Now, $\varphi(\ker\varphi^{i+1})\leq\ker\varphi^i$. Thus, in particular, $$\begin{align*} [\ker\varphi^2:\ker\varphi]&= [\varphi(\ker\varphi^2):\varphi(\ker\varphi)]\\ &=[\varphi(\ker\varphi^2):1]\\ &= |\varphi(\ker\varphi^2)|\\ &\leq |\ker\varphi|. \end{align*}$$ So if $\ker\varphi$ is finite, then so is $\ker(\varphi^2)$. Now proceed by induction to show that $\ker\varphi^r$ is finite.

If $\mathrm{Im}(\varphi)$ has finite index, we can proceed similarly. Note that for arbitrary $H$ and $K$, we know that $\varphi(H) \cong \frac{H}{H\cap\ker\varphi} \cong \frac{H\ker\varphi}{\ker\varphi}$. So $$[G:\mathrm{Im}(\varphi)] \geq [G\ker\varphi:\mathrm{Im}(\varphi)\ker\varphi] = [\varphi(G\ker\varphi):\varphi(\mathrm{Im}(\varphi)\ker\varphi)] = [\mathrm{Im}(\varphi):\mathrm{Im}(\varphi^2)].$$ So the $\mathrm{Im}(\varphi^2)$ also has finite index. Proceed by induction to conclude that $G$ is finite.