For the first: rewrite as
$$x^2 = (y^3 + 12)^2 + 48.$$
The squares differing by 48 are
\begin{align*}
(1,49) \\
(16,64) \\
(121 ,169) \\
\end{align*}
which you can get by solving $(z + 6)^2 = z^2 + 48$, $(z + 4)^2 = z^2 + 48$ and $(z + 2)^2 = z^2 + 48$ respectively.
Choosing $x$ to get the right hand squares should pose no trouble. $(y^3 + 12)^2 = 1$ has no integer solution, $(y^3 + 12)^2 = 16$ has $y = -2$ as a solution and $(y^3 + 12)^2 = 121$
has $y = -1$ as a solution.
So the solution set is
\begin{align*}
(x,y) &= (8, -2) &\lor\\
(x,y) &= (-8, -2) &\lor\\
(x,y) &= (13, -1) &\lor \\
(x,y) &= (-13, -1)
\end{align*}
Hint:
If $5\nmid p$, then, by Fermat's Little Theorem, $p^4\equiv1\pmod5$, so $8p^4-3003\equiv \,?\pmod5$.
Best Answer
If $n=0$ then $m=1$. If $n\geq 1$ then $m\geq 2$ and $3^m-2\equiv 0 \pmod{7}$ iff $m=6k+2$. Moreover $7^n+2\equiv 9\equiv 9^{3k+1}\pmod{36}$ iff $n=6j+1$. Therefore it remains to solve $$3^{6k+2}-7^{6j+1}=2$$ that is $$9(3^{6k}-1)=7(7^{6j}-1)$$ We claim that $k=j=0$ is the only solution and therefore the given equation ${3^m}-{7^n}=2$ has just two solutions $(m,n)$: $(1,0)$ and $(2,1)$.
Assume that $k,j\geq 1$. Note that the prime $43$ divides $(7^6-1)$ and therefore it divides also the right-hand side. It follows that $7$ divides $k$. Then $7^2$ is a factor of $3^{42}-1$ which is a factor of the left-hand side. But then we have a contradiction: $7^2$ divides $7(7^{6j}-1)$ with $j\geq 1$.