Is there positive integer solution of $a^b - c^d = 6$? Question: is there any integer solution of
$$a^b - c^d = e$$
where $e=6$, and $a,b,c,d \geq 2$? Any idea to attack this type of problem? Thanks.
Adam Rubinson comments that $a$ and $c$ are both odd by considering on $\mathbb Z/2 \mathbb Z$. We have $b \neq d$ since otherwise $a-c = 2$ which is impossible because the value of $(a+2)^n - a^n$ increases with $n$ and $a$, and $4^2 - 2^2 = 12 > 6$.

What I've done: for $1 \leq e \leq 9$ and $e \neq 6$, there does exist a solution. But for $e = 6$, I've searched through all integers under $10^{13}$ of the form $a^b$ with no luck. I'll list below the solutions found:
$$
\begin{align}
1 & = 3^2 - 2^3 \\
2 & = 3^3 - 5^2 \\
3 & = 2^7 - 5^3 \\
4 & = 2^3 - 2^2 \\
5 & = 2^5 - 3^3 \\
7 & = 2^4 - 3^2 \\
8 & = (2^3 \times 3 \times 13)^2 - (2 \times 23)^3 \\
9 & = 5^2 - 2^4 \\
\end{align}
$$
 A: THIS IS NOT AN ANSWER, BUT IS TOO LONG FOR THE COMMENTS SECTION
Does $a^b-c^d=6$ have solutions in the integers, $a,b,c,d \ge 2$?
$\{a,c,6\}$ are pairwise coprime, because any prime factor of two of them is a factor of all three of them, and $6$ has only $2,3$ as  prime factors. For $b,d \ge 2$, any prime factor of $a,c$ is present in $a^b,c^d$ at least twice, but $6$ is square-free.
Since $2\not \mid ac$ and $3\not \mid ac$, $a$ and $c$ have the form $6m\pm 1 \Rightarrow a,c \ge 5$. Furthermore, since both $a,c$ are odd, they also can be expressed in the form $4n\pm 1$
In general, $x^{yz}$ can be expressed as $(x^y)^z=X^z$. Accordingly, $b,d$ are coprime; if they share any factor $q$, then the equation can be rewritten $A^q-C^q=6$. In that case, $A>C$, so we can let $A=C+k$, where $C\ge 5, k\ge 2$. There are no $q$th powers of such numbers that differ by less than $24$
For prime number $p$, if $p\mid a$ and $(p-1)\mid d$, or if $p\mid c$ and $(p-1)\mid b$, then $a^b-c^d \equiv \pm 1 \bmod p$. But $6 \equiv \pm 1 \bmod p \iff p \in \{5,7\}$. Thus for $p\ge 11$, $p\mid a \Rightarrow (p-1)\not \mid d$ and $p\mid c \Rightarrow (p-1)\not \mid b$
Unless one of $b,d$ is of the form $2^k$, both $b,d$ will each have at least one odd factor, and we can consider the equation to be expressible in a form with $b,d$ both odd.
For odd exponent $y$, odd number $x$ has the property $x^y\equiv x \bmod 4$ and $x^y\equiv x \bmod 6$. Assuming odd exponents, the equation $a^b-c^d=6$ implies $a \not \equiv c \bmod 4$ and $a \equiv c \bmod 6$. This leads to four permissible cases:
$$a \equiv 1 \bmod 12 \text{ and } c \equiv 7 \bmod 12 \\ 
a \equiv 5 \bmod 12 \text{ and } c \equiv 11 \bmod 12 \\
a \equiv 7 \bmod 12 \text{ and } c \equiv 1 \bmod 12 \\
a \equiv 11 \bmod 12 \text{ and } c \equiv 5 \bmod 12$$
If $2\mid bd$, then the equation can be written either as $A^2-c^d=6$ or $a^b-C^2=6$. Since all bases are of the form $6m \pm 1$, the square term in each equation is $\equiv 1 \bmod 24$ (as is true of the even part $2k$ of the odd exponent $2k+1$) leading to the respective cases
$$A^2 \equiv 1, c\equiv c^d \equiv 19 \bmod 24 \\
a\equiv a^b \equiv 7, C^2 \equiv 1 \bmod 24$$
As a suggestion for a focused search (with no guarantee of success), one might look at the equation $A^4-(5k)^d=6$, since that reduces to $1-0 \equiv 1 \bmod 5$. Bearing in mind that $k$ must be of the form $6m \pm 1$, one could then generate numbers of the form $(30m \pm 5)^d+6$ for various values of $m,d$ and determine whether they have integer fourth roots. Not a high chance  of success, but less work than looking at every number up to $10^{13}$ as OP reports having done.
