Reducing modulo 5 does not make the equations equivalent?

Question

Edit 2: I was trying to solve $2^{n} + 5^{n} - 65 = p^{2}$ for positive integer solutions

$$2^{n} + 5^{n} - 65 \equiv p^{2} \mod 5$$
has different solutions from
$$2^n \equiv p^2 \mod 5$$

Why is this so? Does this mean that both equations are not equivalent? Is reducing modulo $5$ a bad idea for finding integer solutions of $n$ and $p$?

Note: There are more solutions (including the solutions for the previous equation) not shown for $2^n \equiv p^2 \mod 5$, but it still stands that there are suddenly more solutions.

Edit: Equation 2 has too many solutions, including solutions where $n$ is not $0$. Here are some of them:

Answer 1

Here, $5^n \equiv 0 \mod 5$, except for $n = 0$.

Therefore, we have to make a distinction between $n = 0$ and $n \ne 0$.

Moreover, $\phi(5) = 4$.

Therefore,

$$ 2^n \mod 5 \equiv 2^{n \mod 4} \mod 5 $$.

As $n = 0$ is a special case, we have to consider $n = 0, 1, 2, 3, 4$.

Concerning $p$: if $p$ is a solution, then $p+5$ is a solution too. Then, for $p$, we have only to consider its value modulo 5.

At this point, it is easy to consider the different cases.

If $n = 0$:

$$p^2 = 2 \mod 5$$

It is easy to check that $p^2$ is equal to $0, 1, 4$ modulo 5. So, no solution

If $n \ne 0$:

$$p^2 \equiv 2^n \mod 5$$

$$n = 1 \implies p^2=2 \mod 5 \implies no\,solution\,in\,p$$ $$n = 2 \implies p^2=4 \mod 5 \implies p=2\,or\,p=3$$ $$n = 3 \implies p^2=3 \mod 5 \implies no\,solution\,in\,p$$ $$n = 4 \implies p^2=1 \mod 5 \implies p=1\,or\,p=4$$