This problem is wrong. The real problem is: Prove that for all sufficiently large positive integers $d$, at least $99\%$ of the polynomials of the form $\sum_{i\leqslant d}\sum_{j\leqslant d}\pm x^iy^j$are irreducible over the integers.

There is a paper devoted to this problem (https://arxiv.org/pdf/1602.06530.pdf). I follow it here. More details and comments can be found in my blog. Denote the polynomial by $ F(x,y).$ We'll show that the probability of $ F$ being irreducible over $ \mathbb{Z}$ tends to $ 0$ as $ d\to\infty.$ The idea is to eliminate somehow one of the variables by a substitution. Claim. If $ F(x,y)$ is reducible then either $ F(2,y)$ is reducible or $ F(x,2)$ is reducible or $ F(x,y)=f(x)g(y)$. Proof. Suppose $ F(x,y)=f(x,y)g(x,y)$ and $ r$ be the degree of $ y$ in $ F(x,y).$ Putting $ x=2$ we get $$ \displaystyle F(2,y)=f(2,y) g(2,y)$$Note that the coefficient of $ y^r$ in $ F(2,y)$ is odd, that is, it's nonzero. If we assume $ F(2,y)$ is not reducible it means one of the polynomial $ f(2,y)$ or $ g(2,y)$ is a constant and the degree of the other, say $ g(2,y)$ is $ r.$ It means the degree of $ g(x,y)$ with respect to $ y$ is $ r$ hence $ f(x,y)$ is in fact polynomial depending only on $ x$. In the same way, putting $ y=2$ and assuming $ F(x,2)$ is irreducible implies that $ g(x,y)$ depends only on $ y,$ that is, $ F(x,y)=f(x)g(y).\quad \square$ Let us first prove that $ F(x,2),$ resp. $ F(2,y),$ most likely is irreducible. $$ \displaystyle F(x,2)=\sum_{i=0}^d x^i(\pm 2^0\pm 2^1\pm\dots\pm 2^d).$$Note that the expression inside brackets in the right hand side uniquely determines an odd integer in the interval $ [-2^d-1, 2^d+1].$ Let us set $ N:=2^d+2.$ It boils down to prove that the portion of irreducible (over $ \mathbb{Z}$) polynomials of the form $$ \displaystyle F(x)=\sum_{i=0}^d a_i x^i \qquad(1)$$where each $ a_i$ is an odd integer in the interval $ [-N+1,N-1],$ tends to zero when $ d\to\infty.$ Suppose $$ \displaystyle \sum_{i=0}^d a_i x^i = \left( \sum_{i=0}^r b_i x^i\right)\left(\sum_{i=0}^s c_i x^ii \right) \qquad (2)$$where $ r+s=d, r\ge 1,s\ge 1$ and $ b_i,c_i\in\mathbb{Z}.$ Let us consider any polynomial, that can be presented as in $ (2),$ modulo $ N+1.$ Note that the odd numbers in $ [-N+1,N-1]$ different residues modulo $ N+1$ because $ N+1$ is odd. Let us fix $ r,s.$ All possible representation like in $ (2)$ are less than all possible tuples $ (b_0,b_1,\dots,b_r), (c_0,c_1,\dots,c_s)$ of residues modulo $ N+1$ where there are additional conditions for $ b_0,b_r,c_0,c_s.$ All the possible cases for the rest of the coefficients are $ (N+1)^{r-1+s-1}=(N+1)^{d-2}.$ Further, we have $ b_0c_0=a_0, b_rc_s=a_d$ therefore $$ \displaystyle \#\{b_0,c_0,b_r,c_s \}\le \left(\sum_{b_0=1}^N \frac{N}{b_0}\right)\left( \sum_{b_r=1}^N \frac{N}{b_r}\right).$$The right hand side can be estimated by $ C\left(\ln N\right)^2N^2$ for an absolute constant $ C.$ Finally summing over all possible values of $ r,s, r+s=d$ we get that all possible tuples $ (b_i),(c_i)$ for which $ (2)$ holds are less than $$ \displaystyle C d\left(\ln N\right)^2(N+1)^d$$All possible polynomials like in $ (1)$ are $ N^{d+1},$ hence $$ \displaystyle \mathbf{P}( F(x,2) \text{ is irreducible})\le C_1d \frac{\left( \ln N\right)^2}{N}\qquad (3)$$Since $ N=2^d+2,$ it follows $$ \displaystyle \mathbf{P}\big( F(x,2) \text{ is irreducible}\big) \le C_2 \frac{d^3}{2^d}\qquad(4)$$The same estimate holds for $ \mathbf{P}( F(x,2) \text{ is irreducible}).$ It remains to estimate the cases when $$ F(x,y)=f(x)g(y).$$where $ f,g\in \mathbb{Z}[x].$ It easily follows that $ f,g$ are of degree $ d$ with coefficients $ \pm 1.$ Thus $$ \displaystyle \mathbf{P}\big(F(x,y)=f(x)g(y)\big) \le \frac{2^{d+1}2^{d+1}}{2^{(d+1)^2}}\qquad (5)$$Finally, from $ (4)$ and $ (5)$ it yields $$ \displaystyle \lim_{d\to\infty}\mathbf{P}\big(F(x,y) \text{ is irreducible over }\mathbb{Z} \big)=0.$$

Amazing problem. Here is the official solution: let $S{}$ be the set of all polynomials of the form \[\sum_{i\leqslant d}\sum_{j\leqslant d}\pm x^iy^j\]and let $R$ be the subset of $S{}$ consisting of reducible polynomials over the integers. Clearly, $|S| = 2^{(d+1)^2}$, and we will show that $|R|<C'd^32^{d(d+1)}$ for some positive constant $C'{}$. The ratio of $|R|$ to $|S|{}$ is then less than $C'd^32^{-d-1}$, which is arbitrarily small for all large enough $d{}$, whence the conclusion. Before proceeding, observe that no two numbers of the form $\pm1\pm2\pm\cdots\pm2^d$ coincide, and each of them is odd (hence nonzero). Therefore, distinct polynomials $f(x, y)$ in $S{}$ yield distinct degree $d{}$ single variable polynomials $f(2, y)$. A similar statement holds for $f(x, 2)$. Now let $f(x, y)$ be a polynomial in $R$, and write $f(x, y) = g(x, y)h(x, y)$ for some non-constant polynomials with integral coefficients. Let $k$ and $\ell= d-k$ be the $y$-degrees of $g(x, y)$ and $h(x, y)$, respectively, and write \[d =\deg f(2, y) = \deg g(2, y) + \deg h(2, y) \leqslant k + \ell = d,\]to infer that $\deg g(2, y) = k$ and $\deg h(2, y) = \ell$. Hence, either one of $g(x, y)$ and $h(x, y)$ is a polynomial in $x{}$ alone, or $f(2, y)$ is reducible over the integers. Similarly, by considering $x{}$-degrees, either one of $g(x, y)$ and $h(x, y)$ is a polynomial in $y$ alone, or $f(x, 2)$ is reducible over the integers. Consequently, either $f(x, y)$ is separable, say $f(x, y) = g(x)h(y)$, or at least one of the polynomials $f(2, y)$ and $f(x, 2)$ is reducible over the integers. In the former case, both factors are degree $d{}$ polynomials with coefficients $\pm1$. There are at most $2^{d+1}$ sign choices for each factor, so there are at most $2^{2d+2}$ separable polynomials in $R$. To deal with the remaining case, recall that each coefficient of the degree $d$ polynomials $f(2, y)$ and $f(x, 2)$ has the form $\pm1\pm2\pm\cdots\pm2^d$, which is an odd integer of absolute value less than $2^{d+1}$. The lemma below provides an upper bound for the number of non-separable polynomials in $R$. Lemma. The number of degree $d$ single variable polynomials with odd coefficients of absolute value less than $2^{d+1}$, that are reducible over the integers, is less than $Cd^32^{d(d+1)}$ for some absolute constant $C$. Proof. Let $\varphi(t) = a_0 + a_1t + \cdots + a_dt^d$, where the $a_i$ are odd integers of absolute value less than $2^{d+1}$, be reducible over the integers: \[\varphi(t)= a_0 + a_1t + \cdots + a_dt^d=(b_0 + \cdots + b_rt^r)(c_0 + \cdots + c_{d-r}t^{d-r}),\quad\tag{$\dagger$}\]for some positive integer $r < d$ and some integral $b_0, \ldots , b_r, c_0, .\ldots , c_{d-r}$. There are $d-1$ possible choices of $r$; fix $r$. Since $b_0c_0 = a_0$ and $b_rc_{d-r} = a_d$ are both odd integers of absolute value less than $2^{d+1}$, so are $b_0, c_0, b_r$ and $c_{d-r}$. For every odd integer $b_0$ with $|b_0| < 2^{d+1}$, there are less than $2 \cdot 2^{d+1}/|b_0|=2^{d+2}/|b_0|$ choices of $c_0$ so that $b_0c_0$ is an odd integer of absolute value less than $2^{d+1}$. Hence the number of possible choices of the pair $(b_0, c_0)$ is less than \[2\left(\frac{2^{d+2}}{1}+\frac{2^{d+2}}{3}+\cdots+\frac{2^{d+2}}{2^{d+1}-1}\right)<2^{d+3}\sum_{u=1}^{2^{d+1}-1}\frac{1}{u}<2^{d+3}(d+1).\]Similarly, the number of possible choices of the pair $(b_r, c_{d-r})$ is less than $2^{d+3}(d + 1)$. Fix both pairs. Having $r, b_0, c_0, b_r$, and $c_{d-r}$ fixed, each equality of the form $(\dagger)$ can be reduced modulo $2^{d+1} + 1$, and different polynomials $\varphi(t)$ yield different such reductions, as no two odd numbers in the range $[-2^{d+1}, 2^{d+1}]$ are congruent modulo $2^{d+1} + 1$. Therefore, the number of polynomials $\varphi$ satisfying $(\dagger)$ under the above constraints does not exceed the number of choices for $b_i \bmod{2^{d+1} + 1}$ and $c_j \bmod{2^{d+1} + 1}$ with $1\leqslant i\leqslant r-1$ and $1 \leqslant j \leqslant d - r - 1$. This number equals \[(2^{d+1} + 1)^{r-1}(2^{d+1} + 1)^{d-r-1} = (2^{d+1} + 1)^{d-2}.\]All in all, the number of reducible polynomials under consideration does not exceed \[(d-1)2^{2(d+3)}(d+1)^2(2^{d+1} + 1)^{d-2}\leqslant Cd^32^{d(d+1)}\]for some absolute constant $C$. This finishes the proof of the lemma. The above considerations show that $|R|\leqslant 2^{2d+2} + 2\cdot Cd^32^{d(d+1)}\leqslant C'd^32^{d(d+1)}$, as desired.

Cute Problem!