As a complement to Matt's answer, on the intuition: $u[n]$ has value $1$ from $n=0$ on. So basically its energy will increase for ever, because it keeps adding ones for $n\ge0$. Then, you build another signal $x[n]$ that grows way faster because you multiply it with the exponential term $4^n$. 

Therefore, only at the common sense level, one cannot expect it to be an energy signal. It is not feasible. 

And then, goes the proof that the sum of the series actually diverges. Of course, as Matt said, if the  exponential term becomes $4^{-n}$, this geometrical series then decreases fast to zero, and wins over the mere sum of ones in $u[n]$.

From the last edit: always try to use the minimal hypotheses to prove something. You limit the risk of over-abusing results that are not valid (like the geometric series summations when $a\ge 1$), and moreover reduce the risk of mistakes. For instance for the energy part:

$$\sum_{n=-N}^N  (|4^n u[n]|)^2\ge \sum_{n=-N}^N  4^{2n} (u[n])^2\ge \sum_{n=0}^N  |u[n]|^2\ge N+1$$

then $\sum_{n=-N}^N  |4^nu[n]|^2 \to +\infty$. Definitely not an energy signal.