Help finding a less plug-and-chug solution to gross problem. I have the following problem
$$W = + \frac{(3 + 2^{p_{1}})}{2^{(p_1 + p_{2})} - 3^2}$$
Is there a way to prove that if $W$, and all $p_{i}$ are all $\geq 1$ and are integers, that the only solution is $W = 1$ and all $p_{i} = 2$.
I figured that out by just trying values until I got something that gave me an integer W. And it appears that as I make the $p_{i}$ values bigger, the denominator always seems to grow larger faster than the numerator does, hence it will always be a fraction. Thus the only solution can be what I found. But that feels super handwave-y and like there should be a more... real?... way of proving this?
 A: Our equation can be rewritten:  \begin{align}W&=\frac{(3 + 2^{p_{1}})}{2^{(p_1 + p_{2})} - 3^2}\\
 3+2^{p_1}&=W(2^{p_1+p_2}-9)\\
 2^{p_1}&=2^{p_1}2^{p_2}W-9W-3\\
2^{p_1}(2^{p_2}W-1)&=3(3W-1).
\end{align}
Notice here that the left and right sides can both be written as some $2^m\cdot3^n$.  By the fundamental theorem of arithmetic, we must have $$\left\{\begin{align}2^{p_1}&=3W-1\\ 2^{p_2}W-1&=3.\end{align}\right.$$
The second equation is rather easy to solve:  $$2^{p_2}W-1=3\implies2^{p_2}W=4,$$ and since we are given $W,p_2\geq1$, we can only ever have either $(W,p_2)=(1,2)$ or $(2,1)$.
The first equation says that $3W-1$ must be a power of $2$, which means that $3W$ must be odd.  Therefore, $W$ has to be odd, and from the previous conclusion, we now only have $W=1, p_2=2$.  Solving for $p_1$, we have $$3+2^{p_1}=1(2^{p_1+2}-9)\implies12=2^{p_1}(4-1)\implies p_1=2$$ as desired.

Edit: a solution to the problem in the comments:

The equation $$W=\frac{3^2+3\cdot2^{p_1}+2^{p_1+p_2}}{2^{p_1+p_2+p_3}-3^3}$$ can be rewritten as
\begin{align}
2^{p_1+p_2+p_3}W -27W&=9+3\cdot2^{p_1}+2^{p_1+p_2}\\
2^{p_1+p_2+p_3}W-2^{p_1+p_2}&=9+3\cdot2^{p_1}+27W\\
2^{p_1+p_2}(2^{p_3}W-1)&=3(3+2^{p_1}+9W)
\end{align}
Notice that the left has a power of $2$ in $2^{p_1+p_2}$, and so must the right.
Notice that the right has a power of $3$ with maximum exponent $1$, as we would have been able to factor out more factors of $3$ from the expression but alas.
This shows that $2^{p_3}W-1=3$, bringing us halfway to the same conclusion as above.  Solving for $(W,p_3)$ we have $(W,p_3)=(1,2)$ or $(2,1)$. And just like before, since $3+2^{p_3}+9W$ has to be a power of $2$, it has to be even, and as such, $W$ has to be odd. So we have $(W,p_3)=(1,2)$.
Now we can solve for $p_1$ and $p_2$:
\begin{align}
2^{p_1+p_2}&=3+2^{p_1}+9\\
2^{p_1}2^{p_2}-2^{p_1}&=12\\
2^{p_1}(2^{p_2}-1)&=12=2^2\cdot3
\end{align}
It is trivial to see that $p_1=2$ and $p_2=2$.

A: Get rid of the fraction, giving:
$$(2^{p_1+p_2}-9)W= 3 + 2^{p_1} \tag1\label1$$
I claim (A) that the only solution in positive integers is $W=1, p_1=p_2=2$. It's easily shown that this is in fact a solution.
The RHS of ($\ref1$) is always positive. The LHS is positive only if $2^{p_1+p_2}-9$ is positive. Therefore $p_1+p_2 \ge 4$. But if we keep $p_1+p_2=4$ and change $p_1$, the RHS changes without changing the LHS, breaking the equality. Therefore there are no other solutions with $p_1+p_2=4$.
I claim (B) that if $p_1+p_2>4$, the following inequality will always hold:
$$2^{p_1+p_2}-9 > 3 + 2^{p_1} \tag2\label2$$
If this is true, $W < 1$ and is not an integer. We rearrange the inequality:
$$2^{p_1+p_2}-2^{p_1}>12$$
If $p_1+p_2 = 5$, the smallest value the expression on the LHS of ($\ref2$) can take is $2^5-2^4 =16$. If $p_1+p_2=6$, the LHS must be greater than $32$, and so forth.
Since this inequality holds, the claim holds. Hence if $p_1+p_2>4, W <1$, and there are no other solutions to the original equation.
