Show that $11^{n+1}+12^{2n-1}$ is divisible by $133$.
$$11^{(n+1)+1} + 12^{2(n+1)-1} = 11 \cdot 11^{n+1} + 144 \cdot 12^{2n-1} = 11 \left(11^{n+1} + 12^{2n-1}\right) + 133 \cdot 12^{2n-1}$$
- $11 \left(11^{n+1} + 12^{2n-1}\right)$ is divisible by 133 by the induction hypothesis.
- $133 \cdot 12^{2n-1}$ is clearly a multiple of $133$.
In fact, we can prove a stronger result and the proof is easier. The result we will prove is that $$x^2+x+1 \text{ divides }x^{n+1} + (x+1)^{2n-1}$$ for all $n \in \mathbb{N}$. Setting $x=11$ gives the result, you are looking for.
The proof follows immediately from the remainder theorem since $(x^2+x+1) = (x-\omega)(x-\omega^2)$, where $\omega$ is the complex cube-root of unity.
(Remember that $(x-a)$ divides $f(x)$ if and only if $f(x) = 0$)
Plugging in $\omega$ in $x^{n+1} + (x+1)^{2n-1}$ gives us $$\omega^{n+1} + (\omega+1)^{2n-1} = \omega^{n+1} + (-\omega^2)^{2n-1} = \omega^{n+1} - \omega^{4n-2} = \omega^{n+1} - \omega^{3n} \omega^{n-2}\\ =\omega^{n+1} - 1 \times \omega^{n-2} = \omega^{n-2} \left( \omega^3 - 1\right) = 0$$
This gives us that $(x- \omega)$ divides $x^{n+1} + (x+1)^{2n-1}$.
Similarly, plugging in $\omega^2$ in $x^{n+1} + (x+1)^{2n-1}$ gives us $$\omega^{2n+2} + \left( \omega^2 + 1 \right)^{2n-1} = \omega^{2n+2} + (-\omega)^{2n-1} = \omega^{2n+2} - \omega^{2n-1}\\ = \omega^{2n-1} \left( \omega^3 - 1\right) = 0$$ This gives us that $(x- \omega^2)$ divides $x^{n+1} + (x+1)^{2n-1}$.
Hence, $(x-\omega)(x-\omega^2) = x^2 + x + 1$ divides $x^{n+1} + (x+1)^{2n-1}$.
Setting $x=11$ gives us the result you want i.e. $11^2 + 11 + 1 = 133$ divides $11^{n+1} + 12^{2n-1}$.
Note that the above result can also be proved by inducting on $n$.
You had $12^2\cdot 133m-12\cdot11^{n+1}-11^{n+3}$? Then you were very close. Next note that $$-12\cdot11^{n+1}-11^{n+3}=-12\cdot 11^{n+1}-11^2\cdot 11^{n+1}=-(12+11^2)\cdot 11^{n+1}=-133\cdot 11^{n+1}$$ is a multiple of $133$.
$133$ is special here because $133=12^2-11 =11^2+12 = 11\cdot 12+1$. This generalizes to $m^2+m+1 = (m+1)^2-m = m(m+1)+1$, and $m^{n+1}+(m+1)^{2n-1}$ is divisible by $m^2+m+1$ when $m$ is an integer and $n$ is a positive integer.