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Since the coefficients of the polynomial are real numbers, any nonreal roots must come in conjugate pairs. Thus, when we factor $P(z)$ over the integers, each factor is either of the form $z-c,$ where $c$ is an integer, or $$(z-a-bi)(z-a+bi)=z^2-2az+a^2+b^2,$$where $a$ and $b$ are integers, and $b\ne 0.$ Furthermore, the product of the constant terms must be 50, so for each linear factor, $c$ divides 50, and for each quadratic factor, $a^2+b^2$ divides 50. We call these linear and quadratic factors basic factors. For each divisor $d$ of 50, so $d\in \{1,2,5,10,25,50\},$ let $B_d$ be the set of basic factors where the constant term is $\pm d.$

For $d=1,$ any basic quadratic factor must satisfy $$a^2+b^2=1.$$The only solution is $(a,b)=(0,\pm 1),$ which leads to the quadratic factor $z^2+1.$ We also have the linear factors $z\pm 1.$ Hence, $|B_1|=3.$

For $d=2,$ any basic quadratic factor must satisfy $$a^2+b^2=2.$$The solutions are $(a,b)=(\pm 1,\pm 1),$ which leads to the quadratic factors $z^2-2z+2$ and $z^2+2z+2.$ We also have the linear factors $z\pm 2.$ Hence, $|B_2|=4.$

For $d=5,$ the solutions to $$a^2+b^2=5$$are $(a,b)=(\pm 1,\pm 2)$ and $(\pm 2,\pm 1),$ so $|B_5|=6.$

For $d=10,$ the solutions to $$a^2+b^2=10$$are $(a,b)=(\pm 1,\pm 3)$ and $(\pm 3,\pm 1),$ so $|B_{10}|=6.$

For $d=25,$ the solutions to $$a^2+b^2=25$$are $(a,b)=(\pm 3,\pm 4),$ $(\pm 4,\pm 3),$ and $(0,\pm 5),$ so $|B_{25}|=7.$

For $d=50,$ the solutions to $$a^2+b^2=50$$are $(a,b)=(\pm 1,\pm 7),$ $(\pm 5,\pm 5),$ and $(\pm 7,\pm 1),$ so $|B_{50}|=8.$

Now, consider the factors of $P(z)$ which belong in $B_d,$ where $d>1.$ We have the following cases:

$\bullet$ There is one factor in $B_{50}.$

$\bullet$ There is one factor in $B_2,$ and one factor in $B_{25}.$

$\bullet$ There is one factor in $B_5,$ and one factor in $B_{10}.$

$\bullet$ There is one factors in $B_2,$ and two factors in $B_5.$

Case 1: There is one factor in $B_{50}.$

There are 8 ways to choose the factor in $B_{50}.$

Case 2: There is one factor in $B_2,$ and one factor in $B_{25}.$

There are 4 ways to choose the factor in $B_2,$ and 7 ways to choose the factor in $B_{25}.$

Case 3: There is one factor in $B_5,$ and one factor in $B_{10}.$

There are 6 ways to choose the factor in $B_5,$ and 6 ways to choose the factor in $B_{10}.$

Case 4: There is one factors in $B_2,$ and two factors in $B_5.$

There are 4 ways to choose the factor in $B_2,$ and $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$ ways to choose the two factors in $B_5.$

Hence, there are $$8+4\cdot 7+6\cdot 6+4\left(\genfrac{}{}{0ex}{}{6}{2}\right)=132$$ways to choose the factors in $B_d,$ where $d>1.$

After we have chosen these factors, we can include $z+1$ or $z^2+1$ arbitrarily. Finally, the constant coefficient is either 50 or $-50$ at this point. If the coefficient is 50, then we cannot include $z-1.$ If the constant coefficient is $-50,$ then we must include $z-1.$ Thus, whether we include $z-1$ or not is uniquely determined.

Therefore, the total number of polynomials in $G$ is $132\cdot 2^2=\boxed{528}.$