The only solution in natural numbers of the Diophantine equation $(1) \;\; 4^a \cdot 5^b - 3^c \cdot 11^d = 1$, is $(a,b,c,d) = (1,2,2,1)$. Proof: By (1) we have $1 \equiv -3^c \cdot 11^d \equiv 3^c \cdot 3^d = -3^{c+d} \pmod{4}$, yielding $c+d$ is odd. Futhermore by (1) $1 \equiv 4^a \cdot 5^b \equiv 1^a \cdot (-1)^b = (-1)^b \pmod{3}$, which means $b$ is even. Hence $b=2b_1$ for a natural number $b_1$, which combined with the fact that $c+d$ is odd and equation (1) give us $1 = 4^a \cdot 5^b - 3^c \cdot 11^d \equiv 4^a \cdot (5^2)^{b_1} - 3^c \cdot 3^d = 4^a \cdot 1^{b_1} - 3^{c+d} \equiv 4^a - 3 \pmod{8}$, i.e. $4^a \equiv 4 \pmod{8}$, implying $a=1$. Hence by (1) $3^c \cdot 11^d = 4 \cdot 5^b - 1 = 4 \cdot 5^{2b_1} - 1$, or alternatively $(2) \;\; (2 \cdot 5^{b_1} - 1)(2 \cdot 5^{b_1} + 1) = 3^c \cdot 11^d$. Now $GCD(2 \cdot 5^{b_1} - 1, 2 \cdot 5^{b_1} + 1) = 1$ and $1 < 2 \cdot 5^{b_1} - 1 < 2 \cdot 5^{b_1} + 1$ combined with (2) give us $(3) \;\; 2 \cdot 5^{b_1} - 1 = k$, $(4) \;\; {\textstyle 2 \cdot 5^{b_1} + 1 = \frac{3^c \cdot 11^d}{k}}$, where $k \in \{3^c, 11^d\}$. Adding (3) and (4), the result is $(5) \;\; 4 \cdot 5^{b_1} = 3^c + 11^d$. Consequently $0 \equiv 3^c + 11^d \equiv 3^c + 1^d \equiv 3^c + 1 \pmod{5}$, implying $c \equiv 2 \pmod{4}$. Thus, since $c+d$ is odd, $d$ is odd. Assume $k=11^d$. Subtracting (3) from (4), the result is $3^c - 11^d = 2$. Using the fact that $d$ is odd, we find that $2 \equiv -11^d \equiv -(-1)^d = (-1)^{d+1} = 1 \pmod{3}$, a contradiction which means $k \neq 11^d$. Hence $k = 3^c$. Again by subtracting (3) from (4), we obtain $(6) \;\; 11^d - 3^c = 2$. Combining (5) and (6), the result is $(7) \;\; 2 \cdot 5^{b_1} = 3^c + 1$. Assume $b_1 \neq 1$. Then $b_1 \geq 2$, which according to (7) means $25 | 3^c + 1$. Combining the fact $c \equiv 2 \pmod{4}$ and $3^{20} \equiv 1 \pmod{25}$ by Eulers theorem, we obtain $c \equiv 4r+2 \pmod{20}$, where $0 \leq r \leq 4$ and $3^{4r+2} \equiv -1 \pmod{25}$. Checking the five possible values of $r$, we find that $r=2$. In other words, $c = 20q + 10$ for a non-negative integer $q$. Hence $10|c$, yielding $3^{10} - 1 | 3^c - 1$. Now $3^{10} - 1 = (3^5 - 1)(3^5 + 1) = 242 \cdot 244 = 2^3 \cdot 11^2\cdot 61$. Therefore $61 | 3^c - 1$, which according to (6) means $(8) \;\; 11^d = 3^c + 2 \equiv 1 + 2 \equiv 3 \pmod{61}$. We know that $d$ is odd. Hence $11^{\frac{d-1}{2}} = x$ is an integer. Thus by (8) we have $11x^2 \equiv 3 \pmod{61}$, i.e. $(9) \;\; x^2 \equiv 28 \pmod{61}$. Using Gauss' formula $(p/q) \cdot (q/p) = (-1)^{\frac{p-1}{2} \cdot \frac{q-1}{2}}$ ($p,q$ odd primes), we obtain $(28/61) = (2^2 \cdot 7/61) = (7/61) = (61/7) = (12/61) = (2^2 \cdot 3/7) = (3/7) = -(7/3) = -(1/3) = -1$. Therefore the quadratic congruence (9) is not solvable. This means $b_1=1$ (and $b=2b_1=2$), which implies $c=2$ by (7), yielding $d=1$ by (6). Hence the only solution of equation (1) in natural numbers is $(a,b,c,d) = (1,2,2,1)$. q.e.d.

It can be also finished by using mod 121. 9^a+2=11^b. Since 9^a=9,81,3,27,1(mod 121), we obtain that 9^a+2 cannot be divisible by 121. It means that a=b=1.

Here's a very quick approach. Modulo $3$ yields $2^b \equiv 1$, i.e. $b$ is even, say $b=2x$. Now factoring leads to $(2^a \cdot 5^x - 1)(2^a \cdot 5^x + 1) = 3^c \cdot 11^d$. Since the factors on the left are odd and with difference $2$, they are actually coprime (and larger than $1$). Modulo $5$ shows that $2^a \cdot 5^x - 1 = 11^d$ is impossible, hence $2^a \cdot 5^x - 1 = 3^c$ и $2^a \cdot 5^x + 1 = 11^d$. (One can prove that $a$ must be $2$ by mod 8 in the first equation, but do not know if $a=2$ is easily tractable here, anyone?) Now it is sufficient to solve $11^d - 3^c = 2$. Clearly $d=1$, $c=2$ (with $a=1$ and $b=2$) is a solution, let $d\geq 2$. Modulo $121$ we have $3^1 \equiv 3$, $3^2 \equiv 9$, $3^3 \equiv 27$, $3^4 \equiv 81$, $3^5 \equiv 1$ and in particular $3^5 \not\equiv 241 \pmod {121}$. In conclusion, there are no other solutions.

obviously mod 3 -->b even and factorizing+mod 121 kills the problem