$3n+1=a^2$ and $4n+1=b^2$ gives $4a^2-3b^2=1$. With $x=2a$ and $y=b$ we get an instance of Pell's equation.
$$x^2-3y^2=1.$$
Trying small integers, it is obvious that the minimal solution is $4-3=1$, that is $x_1=2$ and $y_1=1$.
Therefore, all the solutions are of the form $(x_k,y_k)$ with $x_k+y_k\sqrt 3=(x_1+y_1\sqrt 3)^k$ or equivalently
$$ \begin{align}
x_{k+1}&=2x_k+3y_k\\
y_{k+1}&=x_k+2y_k,
\end{align} $$
(which already works starting from the trivial solution $x_0=1$, $y_0=0$).
This gives $3y_{k+1} =3x_k+6y_k$ which implies $$\begin{align}&x_{k+2}-2x_{k+1}=3x_k+2x_{k+1}-4x_k\\
\Leftrightarrow &x_{k+2}-4x_{k+1}+x_k=0
\end{align}$$
and $x_0=1$, $x_1=2$.
Clearly, the $x_i$ alternate in sign and we are only interested in the even ones which are those with odd index.
$$\begin{align}x_{2k+1} &= 4x_{2k}-x_{2k-1}\\&=4(4x_{2k-1}-x_{2k-2})-x_{2k-1}\\ &=15x_{2k-1}-x_{2k-1}-x_{2k-3}\\&=14x_{2k-1}-x_{2k-3}.\end{align}$$
Upon division by two, we obtain the recurrence for the solutions $a_k$ to $3n+1=a^2$ subject to the other condition.
$$a_k=14a_{k-1}-a_{k-2},$$
with $a_0=1$ and $a_1=13(=\dfrac{x_3}2)$.
Now, it is trivial to check that modulo $3$, all $a_k$ are $\equiv 1$ because $14\cdot 1 - 1 \equiv 1 \mod 3$. Which implies that $3 \mid a^2-1$.
Furthermore, modulo $7$, all $a_k$ are $\equiv \pm 1$ because $a_k \equiv - a_{k-2} \mod 7$ which implies that $7 \mid a^2-1$.
And finally, modulo $4$, all $a_k$ are $\equiv 1$ because $14-1 \equiv 1 \mod 4$ which implies that $8 \mid (a-1)(a+1)=a^2-1$.
So, in summary, $56 \mid \dfrac{a^2-1}3$ as desired.
Considering squares in $\bmod 9$, we can see that every perfect square must be $0,1,4,7\mod 9$ (these are the quadratic residues $\bmod 9$). However, $10^n+1\equiv 2\mod 9$, so it can never be a square.
Best Answer
To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get $ n^7 + 2^7 = x^2 + 11^2 $ Call $m = n + 2$. Then
$ n^7 + 2^7 = (m - 2)^7 + 2^7 = $
$$\sum_{0 \le k \le 7} \binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = \sum_{1 \le k \le 7} \binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m \sum_{1 \le k \le 7} \binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = m\cdot M.$
Note that $M = m^6 -7\cdot 2\cdot m^5 + - + -21\cdot 2^5m + 7\cdot 2^6$
Hence $\gcd(m,M)$ is a divisor of $7\cdot 2^6$
Note that in $\Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $\Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $\gcd(m,M)$ is either $1$ or $7$.
Observe that in $\Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $\gcd(m,M)=1$
Now if $m \cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $\gcd(m,M) = 1$.
This implies that $m = 1 \pmod 4$ and $n = m - 2 = - 1 . \pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 \pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.