Hint $\ {\rm mod}\,\ 5\!:\,\ k^2 \in \{0,\pm1,\pm2\}^2\equiv \{0, \pm 1\},\,$ none $\equiv 2,\ $ i.e. squares $\not\equiv 2\pmod5$
But by the Rational Root Test, $ $ if $\,x^2\!-15\,$ has a rational root, it is an integer, contra above.
You can't directly use the linked proof because it uses $\,15\mid k^2\,\Rightarrow\, 15\mid k,\,$ which works because $\,15\,$ is squarefree. But $\,5n+2\,$ needn't be squarefree, e.g. it's $\,12\,$ for $\,n=2,\,$ and $\,12\mid 6^2\,$ but $\,12\nmid 6.\,$ But we can always reduce to squarefree radicands: write $\,5n+2 = kj^2.\,$ Then $\sqrt {5n+2} = j\sqrt k\,$ so $\,\sqrt{5n+1}\,$ is rational iff $\sqrt k\,$ is rational. By above $\,5n+2\,$ is not a square so it has squarefree part $\,k>1,\,$ so the linked classical method proves $\sqrt k$ irrational $\,\Rightarrow\sqrt{5n+1} = j\sqrt k\,$ irrational.
For the second, note that if $\ d\mid 5n\!+\!7,\, 3n\!+\!4\ $ then
$${\rm mod}\ d\!:\,\ \dfrac{7}5 \equiv -n \equiv \dfrac{4}3 \,\Rightarrow\, 0\equiv 7\cdot 3 - 5\cdot 4\equiv 1\,\Rightarrow\, d\mid 1$$
Another way is to use Cramer's rule (or elimination) to solve for $\,n\,$ and $\,1\,$ below
$$ \begin{eqnarray} \ 3\,n\, +\, 4\cdot 1 &=& i\\ \\ 5\ n\, +\, \ 7\cdot 1 &\ =\ & j\end{eqnarray}
\quad\Rightarrow\quad \begin{array}\ n \ = \ \ \ \, 7
\,i\, -\, 4\ j \\\\ \color{#c00}{\bf 1}\ =\, {-5}\ \color{#0a0}i\, +\, 3 \ \color{#0a0}j \end{array} $$
Therefore, by the lower equation in the RHS system: $\ n\mid \color{#c00}{\bf 1}\,\ $ if $\,\ n\mid \color{#0a0}{i,\,j}$
Remark $\ $ In the same way we can prove more generally
Theorem $\ $ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \color{#c00}\Delta \gcd(x,y),\ \ \ \color{#c00}{\Delta := {\rm det}\, A}$
e.g. $ $ in OP we have $\,\color{#c00}{\Delta =\bf 1}\,$ so the above yields $\ \gcd(3n+4,5n+7)\mid\color{#c00}{\bf 1}\cdot\gcd(n,1) = 1$