Could you solve $x↑^n2=x↑^m2$? As my title asks, could you solve $x^2=2^x$?
But that's the worrisome part, as I noticed $x↑^n 2=x↑^m 2$ and $2↑^p x=2↑^q x$ will always have a solution at $x=2$.  However, there is bound to be at least another solution.
For example:$$2x=x^2$$has a solution at $x=2,0$.
And $x^2=2^x$ has a solution at $x=2,4,-z$ where $z$ is some hard to determine value.
But I question $2^x=x^x$ having more solutions than just $x=2$.
$x^2=x^x$ appears to have solutions at $x=1,2$ and possibly a negative solution as well.
So what can one say about these solutions other than $x=2$ and some other number?
I've also noticed $2↑^n1=2$ as well.
Which makes my also wonder how odd $2$ is when it comes to higher towers.
And $1$, as that solution arises often.
Due to the confusion in what kind of answer I want, I will make a clarification.$$x↑^12=x\cdot x=x^2$$$$x↑^22=x^x=^2x$$$$x↑^32=^xx=f_2(x)$$$$x↑^42=f_x(x)=g_2(x)$$And so it goes...

As you have probably surmised, $x \uparrow^n 2 = x \uparrow^m 2$ and $2 \uparrow^p x = 2 \uparrow^q x$ have solutions at $x=1$ and $x=2$.  If $m=n$ or $p=q$ then of course any $x$ will be a solution; if $m \neq n$ or $p \neq q$ then there are no other solutions.
First, I will prove the following useful theorems:
Theorem. For $a \ge 2$ and $b \ge 1$
a. $a \uparrow^b c > a\uparrow^b (c-1)$ for $c \ge 2$.
b. $a \uparrow^b c > c$ for $c \ge 1$.
Proof: By double induction on $b,c$.  For $b=1$, we have $a^c = a * a^{c-1} > a^{c-1}$, and it is well known that $a^c \ge 2^c \ge c+1$. (Use the binomial theorem for example.)
For $c=1$, we have $a \uparrow^b 1 = a > 1$.
Assuming the theorem for $b$, and for $b+1$ and $c$, we will prove it for $b+1$ and $c+1$.  Observe that
$$a\uparrow^{b+1}(c+1) = a\uparrow^b (a\uparrow^{b+1}c) > a\uparrow^{b+1} c$$
which proves part a.  For part b, observe that $a\uparrow^{b+1}(c+1) > a\uparrow^{b+1} c > c$, so $a\uparrow^{b+1}(c+1) > c+1$, as desired.
Corollary. Given a fixed $a \ge 2$ and $b \ge 1$, $a \uparrow^b c$ is an increasing function of $c$.
The corollary follows directly from the theorem part a.
Theorem 2. Given $a,c \ge 2$ and $b \ge 1$, $a\uparrow^b c \ge c+2$.
Proof: By induction on $b$.  
If $b=1$, then$a^c \ge 2^c \ge c+2$.
Assume the theorem for $b$.  Then $$a\uparrow^{b+1} c = a\uparrow^b(a\uparrow^{b+1}(c-1)) \ge a\uparrow^b c \ge c+2$$
Now we can address the original equations.  Observe that for $a \ge 2$, $b \ge 1$, $c \ge 3$,
$$a\uparrow^{b+1} c = a\uparrow^b(a\uparrow^{b+1}(c-1) \ge a\uparrow^b(c+1) > a\uparrow^b c$$
so that $a\uparrow^b c$ is an increasing function of $b$; in particular, $2\uparrow^p x =2\uparrow^q x$ has no solutions for $p \neq q$ and $x \ge 3$.
If $a \ge 3$, $b \ge 1$, $c = 2$,
$$a\uparrow^{b+1} 2 = a\uparrow^b a > a\uparrow^b 2$$
so in this case again $a\uparrow^b 2$ is an increasing function of $b$; in particular, $x \uparrow^n 2 = x \uparrow^m 2$ has no solutions for $m \neq n$ and $x \ge 3$.
