$x < 0$: LHS integer iff $x =-1$, but then $LHS = 2 \neq y^{2}$. $(x,y) = (0,2)$ is a solution. for $x = 1,2$ no solution. so assume $x > 2$. LHS is odd, so writing $y = 2n+1$ gives us $2^{x-2}(1+2^{x+1}) = n(n+1)$. $n,n+1$ are coprime, and so are $2^{x-2}, 1+2^{x+1}$. so $n = 2^{x-2}, n+1=1+2^{x+1}$ or vice versa, but both lead to a contradiction

What with $(4,23)$$?$

But $( x,y)=(4,+-23)$ is also a solution

(x,y)=(0,2),(0,-2),(4,23),(4,-23)

as i said about problem 1 i can't believe these are imo problems.i will continue from berti's post since i'm too lazy to write the steps again .it is clear now that $4n/2^{x}$ or $4(n+1)/2^{x}$ are integers. case 1:$4n=a2^{x}\Rightarrow a^{2}*2^{x}+4a=8*2^{x}+4$.it is clear now that $a=1$ or $a=2$.in both cases we get a contradiction. case 2:$4(n+1)=a*2^{x}$again it is easy to see that $a=1$or $a=2$or $a=3$.the only good one is $a=3 \Rightarrow x=4$

This one wasnt so hard and I have solved it but my solution is reallz long but correct

this is really a very straightforward problem..

Say that $x \geq 3$ (the other cases are easily solved). $y^{2}\equiv 1$ is an odd quadratic residue modulo $2^{x}$. I posted a problem some time ago about such odd quadratic residues. ZetaX proved that all residues $\equiv 1 \left( \bmod ~ 8 \right)$ are quadratic residues modulo $2^{x}$ and that they correspond to exactly four numbers. If I remember correctly, you needed to see in which conditions $x_{1}^{2}\equiv x_{2}^{2}$ takes place. Next, $y \equiv 1,-1,2^{x-1}-1 \, \, \textrm{or}\, \, 2^{x-1}+1 \left( \bmod ~ 2^{x}\right)$, so $y \equiv \pm 1 \left( \bmod ~ 2^{x-1}\right)$. Take $y = k \cdot 2^{x-1}\pm 1$, so that the given relation rewrites as $\left( k^{2}-8 \right) 2^{x-2}= 1 \mp k$. Now, we have $k^{2}-8 \geq 1-k$ or $k^{2}-8 \geq 1+k$, except for a small set of numbers, which can be manually checked.

Again easy number theory.

yes again easy number theory problem but this year there are less pefect scorers i think going to be for now I know just three of them also I am not sure please note this : Iurei Boreico Alexandr Magazinov Chinese guy -

Another easy one! My solution: It is easy to show that $x \geq 0$ Now it is obvious for x=0. Now y is odd. Say: $y=2p+1$ where $p \geq 0$ Let’s check for x=1 and 2, and opine that $x\geq 3$ Say: $z=2^{x-2}$ Then, $z+8z^{2}=p(p+1)$ implies, $z(1+8z) = p(p+1)$ it is clear that, $p \geq z$ And one of p or p+1 is even, so: either: $p=kz$ or $p+1=kz$ if $p=kz$ then $z=\frac{1-k}{k^{2}-8}$ and it can be easily shown that, maintaining all the restrictions there is no value for k for which z can be postive. So, $p+1=kz$ for it: $z=\frac{1+k}{k^{2}-8}$ this time, for z to be postive: $k \leq 3$ checking the values k=3 leads z an integer. And it leads to the soution: $(2,23), (2,-23)$

if $x<0 \Rightarrow 1+2^{x}+2^{2x+1}$ integer iff $x=-1 \Rightarrow 1+2^{x}+2^{2x+1}= 2 \neq y^{2}$. if $x=0 \Rightarrow 1+2^{x}+2^{2x+1}=4=(\pm2)^{2}$ if $x>0 \Rightarrow 2^{x}(1+2^{x+1})=(y-1)(y+1) \Rightarrow y=2^{x-1}k+1$ o $y=2^{x-1}k-1$ whit $k>0$ odd. if $y=2^{x-1}k+1 \Rightarrow 1+2^{x+1}=2^{x-2}k^{2}+k \Rightarrow 2^{x-2}=\frac{k-1}{8-k^{2}}$ impossible for all $k$. if $y=2^{x-1}k-1 \Rightarrow 1+2^{x+1}=2^{x-2}k^{2}-k \Rightarrow 2^{x-2}=\frac{k+1}{k^{2}-8}$ impossible for all $k>4$. only solution for $k=3$ and $x=4 \Rightarrow y=\pm23$ the solutons: $(0,\pm2)$ and $(4,\pm23)$ easy problem

i think this was an ugly problem...the short list must have had ugly number theory problems. Last year's number theory problem was very easy too, but in my opinion, nicer.

Hi all. My sol. was similar offical solution; it became different at step $1+m = 2^{x-2}(m^{2}-8)$ (I assumed $y \geq 0$ since $(x,y); (x,-y)$ both are solutions.) Let $2^{x-2}= a$ (then $a > 0$; it's easy to check theres aren't solutions if x<0 so a is integer.) Then $m = ak-1$ where k - positive integer. So $ak = a(a^{2}k^{2}-2ak-7)$; $k(a^{2}k-2a-1)=7$ It means k|7, so k = 1 or 7. If k=7, a isn't integer. If k=1 a = -2 (but $a > 0$) or 4. So k=1, a=4; that gives x=4, y =23 and the final answer is: ${(0;2);(0;-2);(4;23);(4;-23)}$

Just note how easy is to prove there are no solutions for $x>4$ using the binary representation: $2^{x}(1+2^{x+1})=(y-1)(y+1)$ $2^{x}+1 < y < 2^{x+1}$ $y$ wrote as $1b_{2}...b_{k}01...11$ on $x+1$ bits. ($y$-odd, $k \geq 1$) $y-1$: $1b_{2}...b_{k}01...10$ on $x+1$ bits. (one $0$ ending bit) $y+1$: $1b_{2}...b_{k}10...00$ on $x+1$ bits. ($x-k$ of $0$ ending bits) $2^{x}(1+2^{x+1})$: $10...010...0$ on $2x+2$ bits with $x$ of $0$ ending bits. Then $(y-1)(y+1)$ ends both in $x$ and meanwhile in $x-k+1$of $0$ bits so $k=1$: $y-1$: $101...10$ on $x+1$ bits. ($x-3$ of $1$) $y+1$: $110...00$ on $x+1$ bits. $(y-1)(y+1)$: $10001...1010...0$ ($x-4$ of consecutive $1$)

what about $1+a^{x}+a^{2x+1}=y^{2}$??

what about proving that y is a prime number ( in the original imo problem ) ?

(a) lets take $x<0$ : no solutions very easy to state this case. (b) lets take $x=0$ : then we can find a solution. (c) lets take $x>0$ : $2^{x}(1+2^{x+1})=(y-1)(y+1)$ now we know that $2^{x}$ divides RHS and we know that y is odd, so we can write it of the form $2k+1$ so we find that $2^{x-2}(2^{x+1}+1)=k(k+1)$ which means $2^{x-2}$ divides k or k+1, since they are relaively prime. 1. If $2^{x-2}|k$ then we find that k is of the form $2^{p}M$, for $p\geq x-2$ and if $p>x-2$ then we find a simple contradiction. When $p=x-2$ then we find again contradiction. 2. IF $2^{x-2}|k+1$ then we find k of the form $2^{t}N-1$, for $t \geqq-2$ and the case of $t>x-2$ leads to the trivial contradiction. Now lets work on the equality case, we suppose that $t=x-2$. We find that $2^{t+3}+1=2^{t}N^{2}-N$ whic means it is true for $N\geq3$ thus we have $2^{t+3}+1\geq2^{t}9-3=2^{t+3}+2^{t}-3$ which implies to $4 \geq2^{t}$ and from here we find that $2\geqq$ and when we put $t=2$ we find x=4 and also $y=23$ for the other values we have contradictions, so we are done. Answer: $(x,y)={(0,2),(0,-2),(4,23),(4,-23)}$.

Let the equation be $(1+2^{x-1})^{2}+7(2^{2x-2}) = y^{2}$ $7(2^{2x-2}) = (y-1-2^{x-1})(y+1+2^{x-1})$ Then either $y-1-2^{x-1}= 7(2^{k})$, $y+1+2^{x-1}= 2^{2x-2-k}$ or $y+1+2^{x-1}= 7(2^{k})$, $y-1-2^{x-1}= 2^{2x-2-k}$ If $x<0$ there's no solution obviously. For $x=0,1,2,3$ we can check by hand and find $(x,y) = (0,2),(0,-2)$ For $x \ge 4$ and $2x-5 \ge k \ge 3$ we note that taking mod 8, there's contradiction of $y = 1 =-1 (mod 8)$. Thus $k = 0,1,2,2x-2,2x-3,2x-4$. By case checking there is only one solution $(x,y)=(4,23),(4,-23)$

Again, storage from a while ago We claim that the only solutions are $(x,y)=\{(0,\pm2), (4,\pm23)\}$. For simplicity in the future with positive exponents, let $x>2$. It's obvious that if not the only solutions are $(x,y)=(0,\pm2)$. We have that $$2^x(1+2^{x+1})=(y+1)(y-1)\implies 2^{x-1}|y\pm1$$since the two factors differ by 2, meaning one of them has only one power of 2 in it. Case 1: $$y-1=k2^{x-1}\implies k2^{x-1}+2=y+1=2\frac{1+2^{x+1}}{k}\iff k(k2^{x-2}+1)=1+2^{x+1}\iff 2^{x-2}=\frac{k-1}{8-k^2},$$which can be checked to not work for positive integers $k<3$ (otherwise denominator is negative). Case 2: Similar to before, $$y+1=k2^{x-1}\implies k(k2^{x-2}-1)=1+2^{x+1}\iff 2^{x-2}=\frac{k+1}{k^2-8}.$$For this to be an integer, we need numerator greater or equal to denominator, which it is obvious that $k\le 3$. Checking, we find the only solutions are $(k,x,y)=(3,4,\pm23). \blacksquare$

My first ever IMO/ISL NT I think. It is clear that $x$ cannot be negative (To see why taking LHS, when $x$ is negative as a fraction it is obvious that the numerator is odd and the denominator is even, and thus the answer is not an integer, let alone a perfect square). Then we consider the cases when $x=0,1$ for $x=0$ we see that $y=2,-2$ work and there are clearly no solutions for $x=1$. For $x\geq 2$, consider the following factorization; $$1+2^x+2^{2x+1}=y^2\implies 2^x(2^{x+1}+1)=(y-1)(y+1)$$Here, for any power of 2 greater than $2^1$, it is clear that it divides only one of the factors in the RHS. Then proceed based on two cases. Case 1 : $y+1$ is only divisible by $2^1$ (max power of 2). Next, do the following substitution $y-1=2^{x-1}a$ and $y+1=2^{x-1}a+2$. Considering the case when $a=1$, gives us no solutions. Then for $a\geq 2$, notice that; \begin{align*} (y-1)(y+1)&=(2^{x-1}a)(2^{x-1}a+2)\\ 2^x(2^{x+1}+1)&=2^x(a)(2^{x-2}a+1)\\ 2^{x+1}+1&=(a)(2^{x-2}a+1)\\ 2^{x-2}(a^2-8)&= 1-a\\ 2^{x-2}=\frac{1-a}{a^2-8} \end{align*}Now, the LHS is clearly a positive integer, thus the RHS must be so too. Since the numerator is most clearly negative, the denominator must be so too. This means that $a^2\leq 8\implies a\leq 2$, thus the only possibility is $2$. Checking this possibility gives us the solutions $(x,y)=(4,23);(4,-23)$ Case 2 : $y-1$ is only divisible by $2^1$ (max power of 2). Next, do the following substitution $y-1=2^{x-1}a$ and $y+1=2^{x-1}a+2$. \begin{align*} (y-1)(y+1)&=(2^{x-1}a)(2^{x-1}a-2)\\ 2^x(2^{x+1}+1)&=2^x(a)(2^{x-2}a-1)\\ 2^{x+1}+1&=(a)(2^{x-2}a-1)\\ 2^{x-2}(a^2+8)&= a+1\\ 2^{x-2}=\frac{a+1}{a^2+8} \end{align*}It is clear that for all $a\geq 1$, $a+1<a^2+8$, thus the RHS is most clearly not an integer. But the LHS is, which means there are no solutions for this case. Thus, the set of all the solutions is, $(x,y)=(0,2);(0,-2);(4,23);(4,-23)$

We claim the only solutions are $(0, \pm 2)$ and $(4, \pm 23)$. Consider rearranging the equation and factoring like so, $$2^x(2^{x+1} + 1) = (y-1)(y+1)$$ Clearly up to a factor of 2, $y-1$ and $y+1$ are coprime. Moreover $2^x$ and $2^{x+1} + 1$ are coprime. Also due to $\nu_2$ reasons we must have $ x \geq 0$. Thus take cases. Assume that $x > 2$ and note then that either $y - 1$ or $y + 1$ is equal to $2^{x-1}m$ for some $m$ coprime to 2. Let $y - 1 = 2^{x-1}m \iff y = 2^{x-1}m + 1$. Then we have, \begin{align*} 2^x(2^{x+1} + 1) &= (2^{x-1}m)(2^{x-1}m+2)\\ 2^x(2^{x+1} + 1) &= 2^x(2^{x-2}m^2 + m)\\ 2^{x+1} - 2^{x-2}m^2 &= m - 1\\ 2^{x-2}(8 - m^2) &= m - 1\\ 2^{x} &= \frac{4(1-m)}{m^2-8} \end{align*}which has no solutions. Similarly take the second case, $y + 1 = 2^{x-1}m \iff y = 2^{x-1}m - 1$ which gives, \begin{align*} 2^x(2^{x+1} + 1) &= (2^{x-1}m -2)(2^{x-1}m)\\ 2^x(2^{x+1}+1) &= 2^x(2^{x-2}m^2 - m)\\ 2^{x+1} + 1 &= 2^{x-2}m^2-m\\ 2^{x-2}(8 - m^2) &= -1-m\\ 2^{x} &= \frac{4(m+1)}{m^2-8} \end{align*}By inspection $m = 3$ works giving $(4, \pm 23)$. To see that no other values work simply note the denominator grows exponentially, making the RHS a non-integer. Then checking when $x \leq 2$ gives us $(0, \pm 2)$.

We claim the only solutions are $\boxed{(0, \pm 2), (4, \pm 23)}$. Rewrite the equation so that it becomes \[2^x(2^{x+1} + 1) = (y-1)(y+1).\] Up to a factor of $2$, $y-1$ and $y+1$ are coprime and $2^x$ and $2^{x+1}+1$ are also coprime. Clearly, $x \ge 0$, so we will do casework. If $x>2$, note that either $y - 1$ or $y + 1$ is equal to $2^{x-1}m$ for some $m$ coprime to 2. Let $y - 1 = 2^{x-1}m$. Then we have, \begin{align*} 2^x(2^{x+1} + 1) &= (2^{x-1}m)(2^{x-1}m+2)\\ 2^{x+1} - 2^{x-2}m^2 &= m - 1\\ 2^{x} &= \frac{4(1-m)}{m^2-8} \end{align*} which has no solutions. Similarly take the second case, $y + 1 = 2^{x-1}m$ which gives \begin{align*} 2^x(2^{x+1} + 1) &= (2^{x-1}m -2)(2^{x-1}m)\\ 2^{x+1} + 1 &= 2^{x-2}m^2-m\\ 2^{x} &= \frac{4(m+1)}{m^2-8} \end{align*} From inspection, $m = 3$ works giving the pair $(x,y) = (4, \pm 23)$. Otherwise, notice that the denominator grows exponentially, making the right hand side a not an integer. Finally, checking when $x \le 2$ gives us our other case $(x,y) = (0, \pm 2)$.

I claim that the only solutions are $\boxed{(0, \pm 2), (4, \pm 23)}$. Observe first that the cases $x = 0$ and $x < 0$ are trivial, and yield two and no solutions respectively. Henceforth assume $x > 0$. Now observe $y$ is odd, so that $y = 2k + 1$ for $k \in \mathbb{Z}$. Now rearranging, factoring, and simplifying yields \[2^{x - 2}(1 + 2^{x + 1}) = k(k + 1).\]We now split into cases according to when $2^{x - 2} \mid k$ or $2^{x - 2} \mid k + 1$. Case 1: ($2^{x - 2} \mid n$). Then $n = 2^{x - 2}N$ and $n + 1 = 2^{x -2}N + 1$, so that \[2^{x - 2}(1 + 2^{x + 1}) = (2^{x - 2}N)(2^{x - 2}N + 1) \implies 1 + 2^{x + 1} = 2^{x - 2}N^2 + N \implies 2^{x - 2} = \frac{N - 1}{8 - N^2}, \]which has no solutions for $N \in \mathbb{Z}$. Case 2: ($2^{x - 2} \mid n + 1$). Then $n + 1 = 2^{x - 2}N$ and $n = 2^{x - 2} - 1$, so that \[2^{x - 2}(1 + 2^{x + 1}) = (2^{x - 2}N - 1)(2^{x - 2}N) \implies 1 + 2^{x + 1} = N(2^{x - 2}N - 1) \implies 2^{x - 2} = \frac{N + 1}{N^2 - 8} \implies \boxed{N = 3}.\]Hence $\boxed{x = 4}$ and this gives $\boxed{y = \pm 23}$ as the only other solution. $\blacksquare$

$1+2^x+ 2^{2x+1} = y^2$ $=>1+2^x+ 2^{2x}.2 = y^2$ Using Fermat's theorem, $ 2^{2x} \equiv 2 (mod 2x) .... (1) $ Again, $ 2^x \equiv 2 (mod x) $ $ => 2^{2x} \equiv 4 (mod x) ....(2) $ Looking at (1) we can clearly see that the 4 of (2) is easily reducible. Hence, $x \mid 4 => x \leq 4 $ Again, y is an integer so $x \geq -1$ So the bounding we achieve is $ -1 \leq x \leq 4$ Trying these x values gives us, $ (x,y) = (0, \pm2), (4, \pm23) $

The solutions are $(0, \pm 2)$ and $(4, \pm 23)$. If $(x, y)$ is a solution, so is $(x, -y)$, and $y = 0$ clearly does not work, so WLOG assume $y > 0$. We cannot have $x < -1$, since then $0 < 2^x + 2^{2x+1} < 2^{-1} + 2^{-1} = 1$ and $1+2^x+2^{2x+1}$ is not an integer. The cases $x = 0, 1, 2, 3, 4$ can be easily check to yield the solutions $(0, \pm 2)$ and $(4, \pm 23)$. Now assume that $x \geq 5$. From parity, $y$ must be odd. Let $y-1 = 2a$. Then \[ 2^{x-2}(2^{x+1} + 1) = \frac{1}{4}(y-1)(y+1) = a(a+1). \]Since $a$ and $a+1$ are relatively prime, one of them must be a multiple of $2^{x-2}$. Take cases: If $2^{x-2} \mid a$, let $a = k2^{x-2}$. Then we have \[ 2^{x+1} + 1 = 2^{x-2}k^2 + k. \]If $k \geq 3$, then \[ 2^{x-2}k^2 + k \geq 2^{x+1} + 2^{x-2} + 3 > 2^{x+1} + 1, \]so $k<3$. It can be checked that $k=2$ yields the solutions $(0, \pm 2)$ and $k=1$ has no solutions. If $2^{x-2} \mid a+1$, let $a+1 = k2^{x-2}$. Then we have \[ 2^{x+1} + 1 = 2^{x-2}k^2 - k, \]which rearranges to \[ k+1 = 2^{x-2}(k^2 - 8). \]Since we assumed $x \geq 5$, we have \[ k+1 \geq 5(k^2 - 8), \]which implies $k <= 2$. Then the LHS is positive while the LHS is negative, so there are no solutions. Therefore, the only solutions are $(0, \pm 2)$ and $(4, \pm 23)$.

If $x=0$, we get the solutions $(0,2); (0,-2)$ If $x < 0$, let $-x=m \Rightarrow y^{2}=1+\frac{1}{2^{m}}+\frac{1}{2^{2m-1}}$. Then it's easy to conclude that RHS is never an integer. Now, It is trivial that $y$ is an odd integer. So, let $y=2b+1$. This gives us $2^x+2^{2x+1}=4b^2+4b \Rightarrow 2^{x-2}(2^{x+1}+1)=b(b+1)$ Since $b$ is an integer, $b$ or $b+1$ have to be even. Which means $b=2^{x-2}k$ or $b+1=2^{x-2}k$ for some integer $k$. If $b=2^{x-2}k$ then $2^{x-2}=\frac{k-1}{8-k^{2}}$. LHS is positive so RHS also should be. We can easily check that It has no solution. If $b+1=2^{x-2}k$ then $2^{x-2}=\frac{k+1}{k^{2}-8}$. Now this time, LHS is obviously an integer so $k\le 3$. Manually checking gives us the solutions $(4,-23); (4,23)$

The only solutions are $(1,2), (4,23)$, which clearly works. Manually verify $x = 1,2,3,4$ force $y^2 = 4, 37, 137, 529$, which do not result in solutions other than those claimed. Now take $x > 4$. Now we write $2^x(2^{x + 1} + 1) = (y + 1)(y - 1)$, by $\nu_2$ we require $\nu_2 (y + 1) + \nu_2 (y - 1) = x$, and since one of these is forced to be exactly $1$ (both can be zero but that obviously fails), we require the other to be $x - 1$, so we see that $y = 3 \cdot 2^{x - 1} \pm 1$, however this clearly fails by size, the right side is at least $9 \cdot 2^{2x - 2} - 3 \cdot 2^x$, left side is $8 \cdot 2^{2x - 2} + 2^x$, comparing, we desire $9 \cdot 2^{2x - 2} - 3 \cdot 2^x >8 \cdot 2^{2x - 2} + 2^x$, equivalent to $2^{2x - 2} > 2^{x + 2}$, clearly true for $x > 4$, so the right side is always bigger than the left and there are no other solutions. ok after writing up apparently there are negative solutions screw this its the same thing. docks are inevitable..

worst writeup ever concieved LOL We claim that the only solutions are $(0,\pm 2), (4,\pm 23)$. We have $2^x(1+2^{x+1})=y^2-1$. Now, if $y$ is even, then $y^2-1$ is odd so $2^x=1 \implies x=0$, giving us our first solution. If $y$ is odd, then we can substitute $y=2n+1$, giving $$2^x(1+2^{x+1})=(2n+2)(2n) \implies 2^{x-2}(1+2^{x+1})=n(n+1).$$Now, since $n,n+1$ are coprime then $2^{x-2}$ must divide one of them. If it divides $n$, then we can write $n=a \cdot 2^{x-2}$ so we have $a \cdot 2^{x-2} + 1 = 1+2^{x-1}$. Now, it is easy to see by bounding that $a=3$ (and a similar argument can be applied for if it divides $n+1$.) If $2^{x-2}|n$, it gives a negative which is absurd. If $2^{x-2}|n+1$, $n=12$ and $x=4$, which gives $a=\pm23$, so we are done.

For convenience let's suppose $y>0$. If $x\leq 0$ then let $z=-x$ and we have $$\frac{2^z+2}{2^{2z}}=y^2-1\in \mathbb Z \implies z\in \{0,1\}.$$$z=1$ doesn't work and $z=0$ gives us $y=2$. So now consider $x>0$. We can check that $x=1,2,3$ don't work so we take $x\geq 4$. The equation implies that $2^x\mid y^2-1$, so $y$ must be of the form $2^{x-1}t+1$ or $2^{x-1}t-1$. Case 1. $y=2^{x-1}k+1$ ($k\geq 0$): then the equation becomes $2^{x-2}k^2+k=2^{x+1}+1$. If $k\geq 3$ then it's obvious that the $\text{LHS}>\text{RHS}$. So $k=0,1,2$ but we can manually check that none of them work. Case 2. $y=2^{x-1}k-1$ ($k\geq 1$): then the equation becomes $2^{x-2}k^2-2^{x+1}=k+1$. If $k\geq 4$ then the $\text{LHS}>\text{RHS}$. So $k=1,2,3$, but only $k=3$ works. And we find $x=4,y=23$. So the solution set is $$(x,y)\in \{(0,2),(0,-2),(4,23),(4,-23)\}.$$

We have : $2^x(2^{x+1}+1)=(y-1)(y+1)$ Assume $y>0$ Which implies that if $x\neq 0,$ then $ y = 2^{x-1}k\pm1$, where $k$ is a constant. Case $1: +1$ $(2^{x+1}+1)=k\cdot (2^{x-2}k+1)$ if $k\geq 4$, RHS $>$ LHS If $ k = 3$, RHS $>$ LHS If $k =2 $,$2^{x+1}+1 = 2^{x}+2, x = 0 , y = 2$ , $(0,2) , (0,-2)$ If $k \leq 1$, LHS $>$ RHS Case $2: -1$ $(2^{x+1}+1) = k\cdot (2^{x-2}k-1)$ if $k\geq 4, 2^{x+1} + 1 = 2^{x+2} - 4, $ no solution. if $ k = 3 , 2^{x+1} + 1 = 2^{x+1} + 2^{x-2} -3 , x=4, y=23 , (4,23), (4,-23)$ if $k\leq2,$ LHS $>$ RHS, No solution. Our solutions are $(0,\pm2),$ and $(4,\pm23)$

It is easy to check that $x < 0$ does not work. If $x=0,$ we get $(x, y) = (0, \pm 2).$ Thus we now assume that $x>0.$ Then by mod $3,$ and then mod $5$ right after, we find that $x=4k$ for some positive integer $k.$ Then our equation becomes $$2^{4a}(2 \cdot 2^{4a}+1) = (y+1)(y-1).$$WLOG, assume that $y>0.$ Then we can get the bounds $2 \cdot 2^{4a} > y > 2^{4a}+1.$ Note that since $y^2-1 \equiv 2^{4a} \pmod {2^{4a+1}},$ we get that $y \equiv \pm 1 \pmod{2^{4a-1}}.$ Therefore, by the bounds $y=3 \cdot 2^{4a-1} \pm 1,$ and we can test both cases and solve the resulting system of equations, to get $(x, y) = (4, \pm 23).$ Therefore, $(x, y) = (0, \pm 2), (4, \pm 23)$ are the only solutions.

WLOG assume that $x, y$ are positive. Notice that: $(y-1)(y+1) = 2^x(2^{x+1}+1)$. Note that, $y$ must be odd (else , if $y$ is even, LHS would be odd but RHS is even). Note that, $(y-1),(y+1)$ are even and at-least one of them must be divisible by $4$. Thus: $x \ge 3$ as $\nu_2((y-1)(y+1)) = x \ge 3$. Note that: either $\nu_2(y + 1)=x-1$ or $\nu_2(y-1)=x-1$. Thus: $\nu_2(y \pm 1)=x-1$. Plugging $y = 2^{x-1}z \pm 1$ into initial equation: we get: $$1+2^x+2^{2x+1} = (2^{x-1}z \pm 1)^2 \implies 2^{x-2}(z^2-8)=1 \pm z$$Since $z^2-8$ is odd, $\nu_2(1 \pm z) = x-2$. Notice that: $$1+z \ge 1 \pm z =2^{x-2}(z^2-8) \ge 2(z^2-8) \implies 0 \ge 2z^2-z-17.$$Thus: $z=1, 3$. For $z=1$: $\nu_2( 1 + z) = 1$ and $1-z=0$. For $z=3$: $\nu_2(1+z)=2$ and $\nu(1-z)=1$. Thus: $x-2=1, 2$ or $x=3, 4$. Plugging back gives $x=4, y=23$ For $x=0$, we get $y=2$. Note that, there are no solutions for negative values of $x$. Thus, the only solutions are $(0, \pm 2), (4, \pm 23)$.