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Basically I was reading this proof and everything makes sense up to this step. If p is prime number, why does the fact that both 5 and p are prime mean that $5 = (x+1)$ and $p = (x-1)$ or $5 = (x-1)$ and $p = (x+1)$? I know it's correct, I just don't understand why.

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    $5p\ge 10$, so the option of having one of $x+1$ or $x-1$ being equal to $1$ is ruled out. But it's easy to see that either $5|(x+1)$ or $5|(x-1)$ as $5$ is a prime. If $5$ does not equal either of them then it means, say $x+1$ is a multiple of $5$, and if $x+1=5k$ then dividing both sides by $5$ you are left with $p=k(x-1)$ for $k,x-1>1$ which contradicts primality of $p$. Same thing happens with $x-1$ being a multiple of $5$ – 2017-01-26

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Because of the fundamental theorem of arithmetic. The number $5p$ cannot be written in a different way as a product of primes. So in particular, since $5p=(x+1)(x-1)$, and both $x+1$ and $x-1$ are larger than $1$ (if they weren't, we'd have $x=2$, hence $5p=3$, which can't hold), they must be prime and equal to $5$ and $p$, because if they weren't, they'd give a different prime factorization of $5p$.

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    Ok so I mostly get it. When you say "The number $5p$ cannot be written in a different way as a product of primes. Can you give me a example where you can write it a different way (not for $5p$, but maybe $4s$ for some $s$ not necessarily prime. – 2017-01-26
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    You should *explicitly* state that you used the additional hypothesis that $\,x\ge 0.\,$ Without such the claim is false, since $\,x =-4,-6\,$ are solutions too. – 2017-01-26
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    Right. I omitted it, but we are only working with naturals. – 2017-01-26
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    @Chubbles But the question does not state anywhere that you are "only working with naturals". – 2017-01-26
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    @BillDubuque good catch. I just assumed that OP meant natural number solutions, but I agree it'd be better for them to specify in the original post. – 2017-01-26
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    @Chubbles I can take $20=5\cdot 4=10\cdot 2$; the way I write this isn't unique because $4$ wasn't prime in this case. I essentially "moved" one of the $2$s dividing the $4$ over to the $5$ to get $10$. – 2017-01-26
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    $p$ should be greater than $5$ as well, since $25$ is a square and cannot be represented as $(x+1)(x-1)$. – 2017-01-26