Determine $x,y,z$, where $x,y,z \in \mathbb{N}$, such that $2^x+y^2=z!\,$.
I found out that $(0,0,1),(0,1,2),(1,0,2),(1,2,3),(3,4,4)$ are solutions.
I think that for $z \geq5$ there isn't any other solution, but I don't know how to prove this for larger numbers.
I've tried proving it with "the last digit trick", since we know that for any $z \geq5$, $z!$ ends in $0$, but this didn't work for me. And, obviously, $y$ has to be even.
Try factoring $z! = 2^{\alpha_0} 3^{\alpha_1} 5^{\alpha_2} \ldots {p_n}^{\alpha_n}$, where $p_1$ through $p_n$ are the first $n$ odd primes. Then $z! - 2^x = 2^x m$. We know that $\gcd(3^{\alpha_1} 5^{\alpha_2} \ldots {p_n}^{\alpha_n}, m) = 1$, because $\gcd(2^x, 3^{\alpha_1} 5^{\alpha_2} \ldots {p_n}^{\alpha_n}) = 1$ as well.
Okay, sorry, that's not as fruitful as I thought it would be, so I'm just going to elaborate on Wowoju's hint.
Clearly, $z! \equiv 0 \pmod 7$ for all $z > 6$. And $2^x \equiv 1, 2, 4 \pmod 7$. So $z! - 2^x \equiv 6, 5, 3 \pmod 7$. But those are precisely the impossible squares modulo 7.
Now all you have to do is go case-by-case $0 < z < 7$ to check if you have already found all possible solutions.