proof - $(x,y) = (4,6)$ is the only solution for $x^3 + x^2 - 16 = 2^y$ I saw this question and on seeing the answers I believed it did not have to be so complicated. The pair $(x,y) = (4,6)$ only fits the equation.

Find all possible $(x, y)$ pairs for $x^3 + x^2 - 16 = 2^y$.

My efforts:
$$x^3 + x^2 - 16 = 2^y$$
$$x^3 + x^2 = 2^y + 2^4$$
$$x^2(x+1) = 2^4(2^{y-4} +1)$$
$$x^2(x+1) = 4^2(4^{0.5y-2} + 1)$$
Also note that $\gcd(x^2, x+1) = 1$. I think that this result is important.
Now the RHS and LHS look a lot alike. So I presume that there must be a reason to claim that $x$ or $y$ cannot take more than one value. Its adequate to prove that even one of them won't change. I am not able to find a reason why this must happen.
Any help would be appreciated.
 A: $x^2(x+1) = 2^y + 16$
Since LHS is an integer, then we must have $y \ge 0$.
Since RHS is a positive integer, then we must have $x \ge 1$.
$x^2(x+1)$ is strictly increasing for $x \ge 0$.
$2^y + 16$ is strictly increasing for $y \ge 0$.
$$\begin{array}{n|c|c|}
\hline
 n & n^2(n+1) & 2^n + 16 \\ \hline
0 & 0 & 17 \\
1 & 2 & 18 \\
2 & 12 & 20 \\
3 & 36 & 24 \\
4 & 80 & 32 \\
5 & 150 & 48 \\
6 & 252 & 80\\ \hline
\end{array}$$
Note that the table indicates that $(x,y) = (4,6)$ is a solution.
So any solution involving $y \ge 7$ will require $x \ge 5$.
We will show that there is no solution for $y \ge 7$.
So we can assume now  that $x \ge 5$ and $y \ge 7$.
\begin{align}
    x^2(x+1) &= 2^y + 16\\
    x^2(x+1) &= 16(2^{y-4} + 1)\\
\end{align}
Note that $2^{y-4}+1$ must be an odd integer.
So if $x$ is an odd integer, $\gcd(x+1,2^{y-4}+1) = 1$.
$\quad$ Hence $x+1 | 16$
$\quad$ Remembering that $x \ge 5$, we must have $x = 7$ or $x = 15$.
$\quad$Case $1: x = 7$
\begin{align}
  16(2^{y-4}+1) &= 392\\
  2(2^{y-4}+1) &= 49 & \text{Has no solution.}\\
\end{align}
$\quad$Case $2: x = 15$
\begin{align}
  16(2^{y-4}+1) &= 3600\\
  2^{y-4}+1 &= 225 & \\
  2^{y-4} &= 224 \\
  2^{y-4} &= 32 \times 7 & \text{Has no solution.}\\
\end{align}
So if $x$ is an even integer, $\gcd(x^2,2^{y-4}+1) = 1$.
$\quad$ So $x^2 | 16$. 
This can't happen since $x \ge 5$.
A: Why not plot the two function from the LHS and the RHS respectively? You can see the reason immediately and intuitively.
