We can factorize the equation as $(z-2^x)(z+2^x)=3^y$. That is, we can factorize $3^y$ into two powers of $3$ whose difference is $2^{x+1}$, a power of $2$. Thus one of the powers of $3$ must be just $1$. We get then $2^{x+1}=3^y-1$. Then either $x=0$, in which case $y=1$ and $z=2$, or $x\geq 1$ and $2^{x+1}$ is divisible by $4$. Taking the equation $\mod 4$ shows that $y$ is even, say $y=2k$. Then $2^{x+1}=3^{2k}-1=(3^k-1)(3^k+1)$, and so we are able to factorize $2^{x+1}$ into two powers of 2 whose difference is $2$. They must be $2$ and $4$; hence $x=2$ and $k=1$ and $y=2$ and $z=5$. So the two solutions are $(x, y, z)=(0, 1, \pm 2)$ and $(x, y, z)=(2, 2, \pm 5)$.

$4^x\equiv 0\mod 4$ since $x$ is positive integer. Therefore, $z^2\equiv 3^y \mod 4$ which is impossible if $y$ is odd. Thus $y=2y_1$ where $y_1\in \mathbb{Z}^{+}$. This equation is equivalent to $(2^x)^2+(3^{y_1})^2=z^2$. Since $(2^x,3^y)=1$, $2^x=p^2-q^2, 3^{y_1}=2pq$ or $2^x=2pq, 3^{y_1}=p^2-q^2$ by Pythagorean triples. Case $1$ is impossible because $3^{y_1}$ is odd. One of $p,q$ is odd. Therefore, $q=1$ since only odd divisor of $2^x$ is $1$. Thus $3^{y_1}=(p-1)(p+1)$ which implies that $(p-1)=3^k$ and $(p+1)=3^l$ which is if and only if $p=2$. So only solution is $(2,2,5)$.

xeroxia wrote: Find all solutions of the equation $4^x+3^y=z^2$ in positive integers. $3^y = (z - 2^x)(z + 2^x)$ supose $z - 2^x \neq 1$ $\implies$ $3 | 2^{x+1}$ contradiction so $z = 2^x + 1$. So we have $3^y = 2^{x+1} + 1$ let $n = x + 1$. If $n = 1$ then $y = 1$. If $n > 1$ then $y = 2k$ so $2^n = (3^k - 1)(3^k+1)$ but for $n > 1$ $2^{n+1} - 2^n \ge 2$ so $3^k - 1 = 2$ so $k = 1$ and $y = 2$. $(x,y,z) \in \{(0,1,2),(2,2,5)\}$.

So the answer is (x,y,z) {(0,1,+-2),(2,2,+-5)} isn't it

electrovector wrote: So the answer is (x,y,z) {(0,1,+-2),(2,2,+-5)} isn't it In positive integers so (0,1,2) is not an answer.