Find all nonnegative integer solutions to $2^a 3^b 5^c - 7^d 9^e 11^f = 4$.
I think the only solutions are trivial solutions. The tuple $(a,b,c,d,e,f) = (0,1,1,0,0,1)$ is a trivial solution. It could be useful to consider the equation modulo $2,3,4,5,6,7,8,9.$ It could also be useful to find the orders of some numbers modulo some primes. In particular, there's this theorem that states that if $a^b \equiv a^c\mod d$ for some nonzero integer d coprime to a, then $\mathrm{ord}_d(a)$ divides $b-c$.
Considering the equation modulo 2, if $a>0$, then the LHS is odd while the RHS is even, which is a contradiction. Hence $a=0$, so we can ignore the $2^a$ term.
Considering the equation modulo 3, if $b>0,$ then $-7^d 9^e 11^f \equiv 1\mod 3.$ In particular, $e=0$. So $-(-1)^f \equiv 1\mod 3$, which implies that f must be odd. Now if $b=0$, then $(-1)^c - 9^e (-1)^f \equiv 1\mod 3$. In particular, if $e>0$, then $c$ is even. If $e=0$, then c is odd and f is even. Considering the equation modulo 4, $(-1)^b - (-1)^d (-1)^f \equiv 0\mod 4.$ Since $(-1)^x - (-1)^y\in \{-2,2,0\}$ for all x,y, where $-2$ is obtained if $x$ is odd and y is even, 2 is obtained if x is even and y is odd, and 0 is obtained if x and y have the same parity, we know that $b$ and $d+f$ must have the same parity.
Considering the equation modulo 5, if $c>0$, we get $-2^d (-1)^e \equiv -1\mod 5.$ Now the powers of 2 modulo 5 are $2,4,3,1.$ So if $e$ is even, then $d$ is divisible by 4. If $e$ is odd, then $d\equiv 2\mod 4$. If $c=0$, then we get $(-2)^b-2^d (-1)^e \equiv -1\mod 5.$
Considering the equation modulo 6, we get $3^b (-1)^c - 3^e (-1)^f \equiv 4\mod 6.$ But the powers of $3$ are all congruent to $3$ modulo 6, so assuming both b and e are positive, the expression on the LHS must be congruent to one of $-3+3, -3-3, 3+3, 3-3$, all of which equal $0$ modulo $6$. Hence one of $b$ and $e$ has to be zero. If $b=0$, then we get $(-1)^c - 3^e (-1)^f \equiv 4\mod 6,$ so $e=0$ gives $(-1)^c - (-1)^f \equiv -2\mod 6,$ which implies that $c$ is odd and f is even. $e>0$ gives $(-1)^c - 3(-1)^f \equiv -2\mod 6$, so if $f$ is even, then c is even and if f is odd, then $c$ is even. We deduce that $b=0$ or $e=0$.
Considering the equation modulo 7, we get $3^b (-2)^c - 7^d 2^e 4^f \equiv 4\mod 7.$ If $d > 0,$ then $3^b (-2)^c \equiv 4\mod 7$.
Considering the equation modulo 8, we get $3^b (-3)^c - (-1)^d 3^f \equiv 4\mod 8.$ Now the powers of $3$ modulo 8 starting with $3$ are congruent to $3,1.$
Considering the equation modulo 9, we get $3^b 5^c - (-2)^d 9^e 2^f \equiv 4\mod 9.$
A lot of the above congruences seem to lead to complicated equations; only considering the equation modulo 2 and modulo 6 led to particularly simple results.