$6^{(n+2)} + 7^{(2n+1)}$ is divisible by $43$ for $n \ge 1$

$$6^{k+3} + 7^{2k+3} = 6\times6^{k+2} + 49\times7^{2k+1} = 6\times(6^{k+2} + 7^{2k+1}) + 43\times7^{2k+1}$$

Hint $\ $ Note $\ 43 = 6\cdot 7 + 1 = 6^2+6+1\ $ and apply

Lemma $\ \ a^2\!+a+1\mid a^{n+2}+(a+1)^{2n+1}\! =: b$

Proof $\, \ {\rm mod}\,\ a^2\!+a+1\!:\ \color{#0a0}{a(a+1)\equiv -1}$ and $\,\color{#c00}{a^3\equiv 1}\ $ by $\,0\equiv (a\!-\!1)(a^2\!+a+1) = a^3\!-1,\,$ so

$\qquad\quad\! a^{2n+1}b = a^{3n+3} + (\color{#0c0}{a(a+1)})^{2n+1} \equiv\, (\color{#c00}{a^3})^{n+1}\!-1\equiv 0\ $ so $\ b\equiv 0\ \ $ QED

Remark $\ $ If desired you can convert the above to an explicit inductive proof since the proof boils down to $\color{#0a0}{(-1)^{2n+1}\equiv -1}\,$ and $\color{#c00}{1^{n+1}\equiv 1},\,$both of which have obvious inductive proofs (a special case of the Congruence Power Rule inductive proof, see my other answer here).