Notice that if $n$ is even, then $3 \mid m$, but $3 \nmid 5^{n}-3^{n}$, contradiction. So, from now on we assume that $n$ is odd, $n=2k+1$. Obviously $n=1$ is not possible, so $n \geqslant 3$. Notice that $m$ is coprime to $2$, $3$ and $5$.
Let $m_1$ be the smallest positive multiple of $m$ that can be written in the form of either $|5a^{2}-3b^{2}|$ or $|a^{2}-15b^{2}|$ with some integers $a$ and $b$.
Note that $5^{n}-3^{n}=5(5^{k})^{2}-3(3^{k})^{2}$ is a multiple of $m$, so the set of such multiples is non-empty, and therefore $m_1$ is well-defined.

I. First we show that $m_1 \leqslant 5m$. Consider the numbers
$$5^{k+1}x+3^{k+1}y, \quad 0 \leqslant x, y \leqslant \sqrt{m}$$
There are $\lfloor\sqrt{m}\rfloor+1>\sqrt{m}$ choices for $x$ and $y$, so there are more than $m$ possible pairs $(x, y)$. Hence, two of these sums are congruent modulo $m$: $5^{k+1}x_1+3^{k+1}y_1 \equiv 5^{k+1}x_2+3^{k+1}y_2 \pmod{m}$.
Now choose $a=x_1-x_2$ and $b=y_1-y_2$; at least one of $a, b$ is nonzero, and
$$5^{k+1}a+3^{k+1}b \equiv 0 \quad (\bmod m), \quad |a|,|b| \leqslant \sqrt{m}$$
From
$$0 \equiv (5^{k+1}a)^{2}-(3^{k+1}b)^{2}=5^{n+1}a^{2}-3^{n+1}b^{2} \equiv 5 \cdot 3^{n}a^{2}-3^{n+1}b^{2}=3^{n}(5a^{2}-3b^{2}) \pmod{m}$$
we can see that $|5a^{2}-3b^{2}|$ is a multiple of $m$. Since at least one of $a$ and $b$ is nonzero, $5a^{2} \neq 3b^{2}$. Hence, by the choice of $a, b$, we have $0<|5a^{2}-3b^{2}| \leqslant \max(5a^{2}, 3b^{2}) \leqslant 5m$. That shows that $m_1 \leqslant 5m$.

II. Next, we show that $m_1$ cannot be divisible by $2$, $3$ and $5$. Since $m_1$ equals either $|5a^{2}-3b^{2}|$ or $|a^{2}-15b^{2}|$ with some integers $a, b$, we have six cases to check. In all six cases, we will get a contradiction by presenting another multiple of $m$, smaller than $m_1$.
- If $5 \mid m_1$ and $m_1=|5a^{2}-3b^{2}|$, then $5 \mid b$ and $|a^{2}-15(\frac{b}{5})^{2}|=\frac{m_1}{5}<m_1$.
- If $5 \mid m_1$ and $m_1=|a^{2}-15b^{2}|$, then $5 \mid a$ and $|5(\frac{a}{5})^{2}-3b^{2}|=\frac{m_1}{5}<m_1$.
- If $3 \mid m_1$ and $m_1=|5a^{2}-3b^{2}|$, then $3 \mid a$ and $|b^{2}-15(\frac{a}{3})^{2}|=\frac{m_1}{3}<m_1$.
- If $3 \mid m_1$ and $m_1=|a^{2}-15b^{2}|$, then $3 \mid a$ and $|5b^{2}-3(\frac{a}{3})^{2}|=\frac{m_1}{3}<m_1$.
- If $2 \mid m_1$ and $m_1=|5a^{2}-3b^{2}|$, then $|\left(\frac{5a-3b}{2}\right)^{2}-15\left(\frac{a-b}{2}\right)^{2}|=\frac{m_1}{2}<m_1$.
- If $2 \mid m_1$ and $m_1=|a^{2}-15b^{2}|$, then $|5\left(\frac{a-3b}{2}\right)^{2}-3\left(\frac{a-5b}{2}\right)^{2}|=\frac{m_1}{2}<m_1$.
(The last two expressions can be obtained from $(\sqrt{5}a+\sqrt{3}b)(\sqrt{5}-\sqrt{3})=(5a-3b)+\sqrt{15}(b-a)$ and $(a+\sqrt{15}b)(\sqrt{5}-\sqrt{3})=\sqrt{5}(a-3b)+\sqrt{3}(5b-a)$.)
In all six cases, we found that either $\frac{m_1}{2}$, $\frac{m_1}{3}$ or $\frac{m_1}{5}$ is of the form $|5x^{2}-3y^{2}|$ or $|x^{2}-15y^{2}|$. Since $m$ is coprime to $2$, $3$ and $5$, the presented number is a multiple of $m$, but this contradicts the minimality of $m_1$.

III. The last remaining case is $m_1=m$, so either $m=|5a^{2}-3b^{2}|$ or $m=|a^{2}-15b^{2}|$. We will get a contradiction by considering the two sides modulo $3$, $4$ and $5$.
- $2^{n}+65=5a^{2}-3b^{2}$ is not possible, because $2^{n}+65 \equiv 1 \pmod{3}$, but $5a^{2}-3b^{2} \not\equiv 1 \pmod{3}$.
- $2^{n}+65=3b^{2}-5a^{2}$ is not possible, because $2^{n}+65 \equiv 1 \pmod{4}$, but $3b^{2}-5a^{2} \not\equiv 1 \pmod{4}$.
- $2^{n}+65=a^{2}-15b^{2}$ is not possible, because $2^{n}+65 \equiv \pm 2 \pmod{5}$, but $a^{2}-15b^{2} \not\equiv \pm 2 \pmod{5}$.
- $2^{n}+65=15b^{2}-a^{2}$ is not possible, because $2^{n}+65 \equiv 1 \pmod{4}$, but $15b^{2}-a^{2} \not\equiv 1 \pmod{4}$.
We found a contradiction in all cases, that completes the solution.

Suppose again that $5^{n} \equiv 3^{n} \pmod{m=2^{n}+65}$. Like in the first solution, we conclude that $n$ must be odd, and $n \geqslant 3$, so $8 \mid 2^{n}$.
Using Jacobi symbols,
$$-1=\left(\frac{2^{n}+65}{5}\right)=\left(\frac{5}{2^{n}+65}\right)=\left(\frac{5^{n}}{2^{n}+65}\right)=\left(\frac{3^{n}}{2^{n}+65}\right)=\left(\frac{3}{2^{n}+65}\right)=\left(\frac{2^{n}+65}{3}\right)=1,$$
contradiction.