what primes divide $ 1963^{1965}-1963 $

Hint $ $ prime $\ p\mid a^n-a = a(a^{n-1}\!-1)\iff p\mid a\,$ or $\,p\mid a^{n-1}-1.\,$ For the latter, by Fermat

$\!\!\bmod p\!:\ a^{\large 4\cdot 491}\! \equiv 1 \equiv a^{\large p-1}\!\! \iff a^{\large (4\cdot 491,\,p-1)}\equiv1\!$ $\iff\! \bbox[5px,border:1px solid #c00]{a^{\large (4,\,p-1)}\equiv 1}\,$ by $\,\underbrace{491\nmid p\!-\!1}_{\textstyle p\le 491}$

So $\,\color{#90f}{p\mid a(a^4\!-\!1)}=\color{#0a0}{(a\!-\!1)a(a\!+\!1)}\color{#c00}{(a^2\!+\!1)}\,$. In the OP $\,a=1963\,$ and the listed primes (except $7,11$) include all $\rm\color{#90f}{such}$ primes $p\le 491$ except for $61.\,$ Indeed very simple hand calculations show that $\,\bar a = 1963\bmod p \equiv \color{#0a0}{0,\pm1}$ for all listed $p$ except $\,5,7,11$ and of those three only $\,5\,$ has $\,\color{#c00}{\bar a^2\equiv -1}$.

Remark $ $ This yields a closed form for the excluded primes: if $P$ is the product of all listed primes then the excluded primes are those in $\,P/(P,\color{#90f}{a^5\!-a}),\,$ which is $\,77 = 7\cdot 11\,$ here, indeed

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Take $1963$ common, you'll get

$1963(1963^{1964}-1^{1964})$.

Now continuously break to

$1963(1963^{982}+1)(1963^{491}+1)(1963^{491}-1)$.

$1963$ had factors $13, 151$. Also see that unit digit of $(1963^{982}+1)$ is $0$, therefore $2,5$ are also the factors of $z$. Solve further to get other roots

Obviously $z$ inherits all prime factors of $1962=2\times 3^2\times 109$ and $1963=13\times 151$, and the prime factorization $1964=2^2\times491$ also verifies $491$.

Next we'll use Fermat's last theorem. Modulo $5$, $1963=-2$ and $(-2)^{1965}=(-2)^1$ (because $4|1965-1$), so $5|z$. Modulo $7$, $1963=3$ and $3^{1965}=3^3=-1\ne3$, so $7$ fails. Modulo $11$, $1963=5$ and $5^{1965}=5^5=3125=1\ne5$, so $11$ fails.