Divisors of $\left(p^2+1\right)^2$ congruent to $1 \bmod p$, where $p$ is prime 
Let $p>3$ be a prime number.  How to prove that $\left(p^2+1\right)^2$ has no divisors congruent to $1 \bmod p$, except the trivial ones $1$, $p^2+1$, and $\left(p^2+1\right)^2$? 

When $p=3$, you also have $p+1$ as a divisor of $\left(p^2+1\right)^2$.

Assume there are factors of $\def\num{(p^2+1)^2}\num$ which are congruent to $1$ modulo $p$. Then both the factor and the quotient are $\equiv1\pmod p$ as $\num\equiv1\pmod p$, so we can write $(np+1)(mp+1)=\num$.
Then
$$
(nmp+n+m)p+1=p^4+2p^2+1,
$$
so $nmp+n+m=p^3+2p$. This shows that $p$ divides $n+m$.
Without loss of generality, write $n=a_np+k$ and $m=a_mp-k$ where $a_m\geq1$ and $0\leq k<p$.
Substituting in the equation $nmp+n+m=p^3+2p$, we obtain
$$a_na_mp^2+(a_m-a_n)kp-k^2+a_n+a_m=p^2+2.$$
Now $(np+1)(mp+1)=(p^2+1)^2$ means one of $n$ and $m$ is $\leq p$, otherwise $(np+1)(mp+1)>\num$. This implies we only have to consider three cases: $a_n=1,\,k=0$, $a_n=0$ or $a_m=1$.
$a_n=1,\,k=0$:
The equation becomes $$a_mp^2+a_m=p^2+1.$$ This means $a_m=1$, and we obtain the solution $(p^2+1)(p^2+1)=(p^2+1)^2$.
$a_n=0$:
The equation becomes $$a_mkp-k^2+a_m=p^2+2.$$
If $a_m>p$, then the left-hand side is larger than the right-hand side of the equation, so $a_m\leq p$.
Since $a_m-k^2\leq p$, we must have $a_m=p$ and $k=1$. It follows that $a_m-k^2=2$, so $p=3$. And we obtain the case $(p+1)((p^2-1)p+1)=(p^2+1)^2$ for $p=3$.
$a_m=1$:
Write the equation as $$((a_np+k)p+1)((p-k)p+1)=p^4+2p^2+1.$$
Hence $$(a_np+k)(p-k)p^2+(a_np+p)p+1=p^4+2p^2+1$$ and $$(a_np+k)(p-k)+a_n=p^2+1.$$
Namely, $$a_n(p^2-kp+1)=p^2+1-kp+k^2.$$
This means that $p(p-k)+1$ divides $k^2$. Note that here $k\ne0$, otherwise it is just the case $(p^2+1)(p^2+1)=(p^2+1)^2$. Substituting $k$ by $p-k$, we obtain $pk+1\mid(p-k)^2$.
By this question, this is impossible.
Therefore the factors of $\num$ which are $\equiv1\pmod p$ are as described.

Hope this helps. Perhaps I shall post this as community wiki, as it involves an answer from other users.
